26. Inverse Circular Functions Solutions#

  1. Let tan⁡−1(−1)=θ\tan^{-1}(-1) = \theta then tan⁡θ=−1.\tan\theta = -1. Also, −π2<θ<π2-\frac{\pi}{2}< \theta < \frac{\pi}{2}

    The only value in this range which satisfied this equation is −π4.-\frac{\pi}{4}.

  2. Let cot⁡−1(−1)=θ,\cot^{-1}(-1) = \theta, then cot⁡θ=−1\cot\theta = -1 and 0<θ<π0<\theta<\pi

    ⇒θ=3π4\Rightarrow \theta = \frac{3\pi}{4}

  3. Let sin⁡−1(−32)=θ\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \theta then sin⁡θ=−32\sin\theta = -\frac{\sqrt{3}}{2}

    Also, −π2≤θ≤π2⇒θ=−π3-\frac{\pi}{2}\leq \theta\leq \frac{\pi}{2} \Rightarrow \theta = -\frac{\pi}{3}

  4. sin⁡[π3−sin⁡−1(−12)]\sin\left[\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2}\right)\right]

    =sin⁡[π3−{−sin⁡−112}]=\sin\left[\frac{\pi}{3} -\{-\sin^{-1}\frac{1}{2}\}\right]

    =sin⁡[π3+π6]=1= \sin\left[\frac{\pi}{3} + \frac{\pi}{6}\right] = 1

  5. sin⁡[cos⁡−1(−12)]=sin⁡[π−cos⁡−112]\sin\left[\cos^{-1}\left(-\frac{1}{2}\right)\right] = \sin\left[\pi - \cos^{-1}\frac{1}{2}\right]

    =sin⁡2π3=32= \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}

  6. sin⁡[tan⁡−1(−3)+cos⁡−1−32]\sin\left[\tan^{-1}(-\sqrt{3}) + \cos^{-1}\frac{-\sqrt{3}}{2}\right]

    =sin⁡[−π3+π−π6]= \sin\left[-\frac{\pi}{3} + \pi- \frac{\pi}{6}\right]

    =sin⁡π2=1= \sin\frac{\pi}{2} = 1

  7. Given expression is tan⁡[12cos⁡−153]\tan\left[\frac{1}{2}\cos^{-1}\frac{\sqrt{5}}{3}\right]

    Let cos⁡−153=2θ\cos^{-1}\frac{\sqrt{5}}{3} = 2\theta then cos⁡2θ=53\cos2\theta = \frac{\sqrt{5}}{3}

    ⇒1−tan⁡2θ1+tan⁡2θ=53\Rightarrow \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{\sqrt{5}}{3}

    By componendo and dividendo 2tan⁡2θ2=3−53+5\frac{2\tan^2\theta}{2} = \frac{3 - \sqrt{5}}{3 + \sqrt{5}}

    ⇒tan⁡θ=±(3−52)\Rightarrow \tan\theta = \pm\left(\frac{3 - \sqrt{5}}{2}\right)

    Given 0≤2θ≤π⇒0≤θ≤π20 \leq 2\theta \leq \pi \Rightarrow 0\leq \theta \leq \frac{\pi}{2}

    i.e. θ\theta lies in first quadrant. ⇒tan⁡θ=3−52\Rightarrow \tan\theta = \frac{3 - \sqrt{5}}{2}

  8. Given expression is sin⁡−1(sin⁡2π3)\sin^{-1}\left(\sin\frac{2\pi}{3}\right)

    Let sin⁡−1(sin⁡2π3)=θ\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \theta

    then sin⁡θ=sin⁡2π3\sin\theta = \sin\frac{2\pi}{3} and −π2≤θ≤π2-\frac{\pi}{2}\leq\theta\leq \frac{\pi}{2}

    =sin⁡(π−π3)=sin⁡π3⇒θ=π3=\sin\left(\pi - \frac{\pi}{3}\right) = \sin\frac{\pi}{3} \Rightarrow \theta = \frac{\pi}{3}

    sin⁡−1(sin⁡2π3)=π3\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \frac{\pi}{3}

  9. Let sin⁡−132=θ⇒sin⁡θ=32\sin^{-1}\frac{\sqrt{3}}{2} = \theta \Rightarrow \sin\theta = \frac{\sqrt{3}}{2}

    ⇒θ=π3\Rightarrow \theta = \frac{\pi}{3} −π2≤θ≤π2-\frac{\pi}{2}\leq\theta\leq \frac{\pi}{2}

  10. Let tan⁡−1−13=θ\tan^{-1}\frac{-1}{\sqrt{3}} =\theta

    then tan⁡θ=−13=tan⁡(−π3)\tan\theta = \frac{-1}{\sqrt{3}} =\tan\left(\frac{-\pi}{3}\right)

    ⇒θ=−π3\Rightarrow \theta = -\frac{\pi}{3}

  11. Let cot⁡−1(−3)=θ⇒cot⁡θ=−3\cot^{-1}(-\sqrt{3}) = \theta \Rightarrow \cot\theta = -\sqrt{3}

    cot⁡θ=cot⁡(π−π6)\cot\theta = \cot\left(\pi - \frac{\pi}{6}\right)

    θ=5π6\theta = \frac{5\pi}{6}

  12. Let cot⁡−1cot⁡5π4=θ⇒cot⁡θ=cot⁡(π+π4)\cot^{-1}\cot\frac{5\pi}{4} = \theta \Rightarrow \cot\theta = \cot\left(\pi + \frac{\pi}{4}\right)

    cot⁡θ=cot⁡π4⇒θ=π4\cot\theta = \cot\frac{\pi}{4} \Rightarrow \theta = \frac{\pi}{4}

  13. Let tan⁡−1(tan⁡3π4)=θ\tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \theta

    ⇒tan⁡θ=tan⁡3π4=tan⁡(π−π4)\Rightarrow \tan\theta = \tan\frac{3\pi}{4} = \tan\left(\pi - \frac{\pi}{4}\right)

    ⇒θ=−π4\Rightarrow \theta = -\frac{\pi}{4}

  14. sin⁡−112+cos⁡−112=π6+π3=π2\sin^{-1}\frac{1}{2} + \cos^{-1}\frac{1}{2} = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}

  15. Let tan⁡−1=θ⇒tan⁡θ=34⇒cos⁡θ=45\tan^{-1} = \theta \Rightarrow \tan\theta = \frac{3}{4} \Rightarrow \cos\theta = \frac{4}{5}

  16. cos⁡[cos⁡−1(32)+π6]\cos\left[\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) + \frac{\pi}{6}\right]

    =cos⁡[π6+π6]=cos⁡π3=12=\cos\left[\frac{\pi}{6} + \frac{\pi}{6}\right] = \cos \frac{\pi}{3} = \frac{1}{2}

  17. L.H.S. =2tan⁡−113+tan⁡−117= 2\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7}

    =tan⁡−12∗131−19+tan⁡−117= \tan^{-1}\frac{2*\frac{1}{3}}{1 - \frac{1}{9}} + \tan^{-1}\frac{1}{7}

    =tan⁡−134+tan⁡−117= \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{1}{7}

    =tan⁡−134+171−328=tan⁡−11=π4== \tan^{-1}\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{28}} = \tan^{-1}1 = \frac{\pi}{4} = R.H.S.

  18. L.H.S. =tan⁡−113+tan⁡−117+tan⁡−115+tan⁡−118= \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8}

    =tan⁡−113+171−121+tan⁡−115+181−140= \tan^{-1}\frac{\frac{1}{3} + \frac{1}{7}}{1 - \frac{1}{21}} + \tan^{-1}\frac{\frac{1}{5} + \frac{1}{8}}{1 - \frac{1}{40}}

    =tan⁡−112+tan⁡−113=tan⁡−11=π4== \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}1 = \frac{\pi}{4} = R.H.S.

  19. L.H.S. =sin⁡−145+sin⁡−1513+sin⁡−11665= \sin^{-1}\frac{4}{5} + \sin^{-1}\frac{5}{13} + \sin^{-1}\frac{16}{65}

    =sin⁡−1(451−25169+5131−1625)+sin⁡−11665= \sin^{-1}\left(\frac{4}{5}\sqrt{1 - \frac{25}{169}} + \frac{5}{13}\sqrt{1 - \frac{16}{25}}\right) + \sin^{-1}\frac{16}{65}

    =sin⁡−1(4865+1565)+sin⁡−11665=sin⁡−16465+sin⁡−11665= \sin^{-1}\left(\frac{48}{65} + \frac{15}{65}\right) + \sin^{-1}\frac{16}{65} = \sin^{-1}\frac{64}{65} + \sin^{-1}\frac{16}{65}

    =sin⁡−1(6365.1−162652+16651−632652)= \sin^{-1}\left(\frac{63}{65}.\sqrt{1 - \frac{16^2}{65^2}} + \frac{16}{65}\sqrt{1 - \frac{63^2}{65^2}}\right)

    sin⁡−1632+162652=sin⁡−1=π2=\sin^{-1}\frac{63^2 + 16^2}{65^2} = \sin^{-1} = \frac{\pi}{2} = R.H.S.

  20. We have to prove that 4tan⁡−115−tan⁡−1170+tan⁡−1199=π44\tan^{-1}\frac{1}{5} - \tan^{-1}\frac{1}{70} + \tan^{-1}\frac{1}{99} = \frac{\pi}{4}

    ⇒4tan⁡−115=π4+tan⁡−1170−tan−1199\Rightarrow 4\tan^{-1}\frac{1}{5} = \frac{\pi}{4} + \tan^{-1}\frac{1}{70} - tan^{-1}\frac{1}{99}

    L.H.S. =4tan⁡−115=2tan⁡−12.151−125=2tan⁡−1512= 4\tan^{-1}\frac{1}{5} = 2\tan^{-1}\frac{2.\frac{1}{5}}{1 - \frac{1}{25}} = 2\tan^{-1}\frac{5}{12}

    =tan⁡−12.5121−25144=tan⁡−1120119= \tan^{-1}\frac{2.\frac{5}{12}}{1 - \frac{25}{144}} = \tan^{-1}\frac{120}{119}

    tan⁡−1170−tan⁡−1199=tan⁡−1170−1991+170.199\tan^{-1}\frac{1}{70} - \tan^{-1}\frac{1}{99} = \tan^{-1}\frac{\frac{1}{70} - \frac{1}{99}}{1 + \frac{1}{70}.\frac{1}{99}}

    =tan⁡−11239= \tan^{-1}\frac{1}{239}

    R.H.S. =tan⁡−11+tan⁡−11239=tan⁡−1120119= \tan^{-1}1 + \tan^{-1}\frac{1}{239} = \tan^{-1}\frac{120}{119}

    Thus, L.H.S. = R.H.S.

  21. We have to prove that cot⁡−19+cosec⁡−1414=π4\cot^{-1}9 + \cosec^{-1}\frac{\sqrt{41}}{4} = \frac{\pi}{4}

    cot⁡−19=tan⁡−119\cot^{-1}9 = \tan^{-1}\frac{1}{9}

    Let cosec⁡−1414=θ⇒cosec⁡θ=414\cosec^{-1}\frac{\sqrt{41}}{4} = \theta \Rightarrow \cosec\theta = \frac{\sqrt{41}}{4}

    Since we have to consider principal values only −π2≤θ≤π2-\frac{\pi}{2}\leq \theta\leq \frac{\pi}{2} and θ≠0\theta \neq 0

    As cosec⁡θ\cosec\theta is +ve here, θ\theta lies between 00 and π/2,\pi/2, hence tan⁡θ\tan\theta must be positive.

    ⇒tan⁡θ=45\Rightarrow \tan\theta = \frac{4}{5}

    tan⁡−119+tan⁡−145=tan⁡−119+451−1945\tan^{-1}\frac{1}{9} + \tan^{-1}\frac{4}{5} = \tan^{-1}\frac{\frac{1}{9} + \frac{4}{5}}{1 - \frac{1}{9}\frac{4}{5}}

    =tan⁡−14145.4541=tan⁡−11=π4==\tan^{-1}\frac{41}{45}.\frac{45}{41} = \tan^{-1}1 = \frac{\pi}{4} = R.H.S.

  22. We have to prove that 4(cot⁡−13+cosec⁡−15)=π4(\cot^{-1}3 + \cosec^{-1}\sqrt{5}) = \pi

    cot⁡−13=tan⁡−113\cot^{-1}3 = \tan^{-1}\frac{1}{3}

    cosec⁡−15=tan⁡−112\cosec^{-1}\sqrt{5} = \tan^{-1}\frac{1}{2}

    L.H.S. =4(tan⁡−113+tan⁡−112)=4(tan⁡−112+131−12.13)=4(\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{2}) = 4\left(\tan^{-1}\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2}.\frac{1}{3}}\right)

    =4tan⁡−11=π== 4\tan^{-1}1 = \pi = R.H.S.

  23. We have to prove that tan⁡−1x=2tan⁡−1[cosec⁡tan⁡−1x−tan⁡cot⁡−1x]\tan^{-1}x = 2\tan^{-1}[\cosec\tan^{-1}x - \tan\cot^{-1}x]

    R.H.S. =2tan⁡−1[cosec⁡cosec⁡−11+x2x−tan⁡tan⁡−11x]= 2\tan^{-1}[\cosec\cosec^{-1}\frac{\sqrt{1 + x^2}}{x} - \tan\tan^{-1}\frac{1}{x}]

    =2tan⁡−1(1+x2−1x)= 2\tan^{-1}\left(\frac{\sqrt{1 + x^2} - 1}{x}\right)

    Let x=tan⁡θ,x = \tan\theta, then R.H.S. =2tan⁡−1(secθ−1tan⁡θ)= 2\tan^{-1}\left(\frac{sec\theta - 1}{\tan\theta}\right)

    =2tan⁡−1(1−cos⁡θsin⁡θ)=2tan⁡−1(2sin⁡2θ22sin⁡θ2cos⁡θ2)=2\tan^{-1}\left(\frac{1 - \cos\theta}{\sin\theta}\right) = 2\tan^{-1}\left(\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right)

    =2tan⁡−1.tan⁡θ2=θ=tan−1x== 2\tan^{-1}.\tan\frac{\theta}{2} = \theta = tan^{-1}x = L.H.S.

  24. ∵0<b≤a∴a−ba+b\because 0 < b \leq a \therefore \sqrt{\frac{a - b}{a + b}} is real.

    L.H.S. =2tan⁡−1[a−ba+btan⁡x2]= 2\tan^{-1}\left[\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}\right]

    =cos⁡−1[1−a−ba+btan⁡2x21+a−ba+btan⁡2x2]= \cos^{-1}\left[\frac{1 - \frac{a- b}{a + b}\tan^2\frac{x}{2}}{1 + \frac{a - b}{a + b}\tan^2\frac{x}{2}}\right]

    =cos⁡−1[a(1−tan⁡2x2)+b(1+tan⁡2x2)a(1+tan⁡2x2)+b(1+tan⁡2x2)]= \cos^{-1}\left[\frac{a\left(1 - \tan^2\frac{x}{2}\right)+ b\left(1 + \tan^2\frac{x}{2}\right)}{a\left(1 + \tan^2\frac{x}{2}\right)+ b\left(1 + \tan^2\frac{x}{2}\right)}\right]

    =cos⁡−1[a(1−tan⁡2x21+tan⁡2x2+b)a+b1−tan⁡2x21+tan⁡2x2]= \cos^{-1}\left[\frac{a\left(\frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2} + b}\right)}{a + b\frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}}}\right]

    =cos⁡−1[b+acos⁡xa+bcos⁡x]== \cos^{-1}\left[\frac{b + a\cos x}{a + b\cos x}\right] = R.H.S.

  25. L.H.S =tan⁡−1x−y1+xy+tan⁡−1y−z1+yz+tan⁡−1z−x1+zx= \tan^{-1}\frac{x - y}{1 + xy} + \tan^{-1}\frac{y - z}{1 + yz} + \tan^{-1}\frac{z - x}{1 + zx}

    =tan⁡−1x−tan⁡−1y+tan⁡−1y−tan⁡−1z+tan⁡−1z−tan⁡−1x=0= \tan^{-1}x - \tan^{-1}y + \tan^{-1}y - \tan^{-1}z + \tan^{-1}z - \tan^{-1}x = 0

    R.H.S. =tan⁡−1(x2−y21+x2y2)+tan⁡−1(y2−z21+y2z2)+tan⁡−1(z2−x21+z2x2)= \tan^{-1}\left(\frac{x^2 - y^2}{1 + x^2y^2}\right) + \tan^{-1}\left(\frac{y^2 - z^2}{1 + y^2z^2}\right) + \tan^{-1}\left(\frac{z^2 - x^2}{1 + z^2x^2}\right)

    =tan⁡−1x2−tan⁡−1y2+tan⁡−1y2−tan⁡−1z2+tan⁡−1z2−tan⁡−1x2=0= \tan^{-1}x^2 - \tan^{-1}y^2 + \tan^{-1}y^2 - \tan^{-1}z^2 + \tan^{-1}z^2 - \tan^{-1}x^2 = 0

    ∴\therefore L.H.S. = R.H.S.

  26. We have to prove that sin⁡cot⁡−1tan⁡cos⁡−1x=x\sin\cot^{-1}\tan\cos^{-1}x = x

    L.H.S. =sin⁡cot⁡−1tan⁡tan⁡−11−x2x= \sin\cot^{-1}\tan\tan^{-1}\frac{\sqrt{1 - x^2}}{x}

    =sin⁡cot⁡−11−x2x= \sin\cot^{-1}\frac{\sqrt{1 - x^2}}{x}

    Let cot⁡−11−x2x=θ\cot^{-1}\frac{1 - x^2}{x} = \theta then cot⁡θ=1−x2x\cot\theta = \frac{\sqrt{1 - x^2}}{x}

    sin⁡θ=x\sin\theta = x

    Thus, L.H.S. = R.H.S.

  27. Case I: When π4<x<π2\frac{\pi}{4}<x<\frac{\pi}{2}

    0<cot⁡x<10<\cot x< 1 and 0<cot⁡3x<1∴0<cot⁡xcot⁡3x<10<\cot^3x<1 \therefore 0<\cot x\cot^3x<1

    tan⁡−1cot⁡x+tan⁡−1cot⁡3x=tan⁡−1cot⁡x+cot⁡3x1−cot⁡xcot⁡3x\tan^{-1}\cot x + \tan^{-1}\cot^3x =\tan^{-1}\frac{\cot x + \cot^3x}{1 - \cot x\cot^3x}

    =tan⁡−1cot⁡x1−cot⁡2x=tan⁡−1tan⁡xtan⁡2−1=\tan^{-1}\frac{\cot x}{1 - \cot^2x} = \tan^{-1}\frac{\tan x}{\tan^2 - 1}

    =−tan⁡−1(12tan⁡2x)= -\tan^{-1}\left(\frac{1}{2}\tan 2x\right)

    ⇒tan⁡−1(12tan⁡2x)+tan⁡−1cot⁡x+tan⁡−1cot⁡3x=0\Rightarrow \tan^{-1}\left(\frac{1}{2}\tan 2x\right) + \tan^{-1}\cot x + \tan^{-1}\cot^3x = 0

    Case II: When 0<x<π40<x<\frac{\pi}{4}

    cot⁡x>1\cot x> 1 and cot⁡3x>1\cot^3x > 1

    ⇒tan⁡−1cot⁡x+tan⁡−1cot⁡3x=π−tan⁡−1(12tan⁡2x)\Rightarrow \tan^{-1}\cot x + \tan^{-1}\cot^3x = \pi - \tan^{-1}\left(\frac{1}{2}\tan2x\right)

    ⇒tan⁡−1(12tan⁡2x)+tan⁡−1cot⁡x+tan⁡−1cot⁡3x=π\Rightarrow \tan^{-1}\left(\frac{1}{2}\tan 2x\right) + \tan^{-1}\cot x + \tan^{-1}\cot^3x = \pi

  28. tan⁡−112+tan⁡−113=tan⁡−112+131−12.13\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2}.\frac{1}{3}}

    =tan⁡−15/65/6=tan⁡−11=π4= \tan^{-1}\frac{5/6}{5/6} = \tan^{-1}1 = \frac{\pi}{4}

    tan⁡−135+tan⁡−114=tan⁡−135+141−35.14\tan^{-1}\frac{3}{5} + \tan^{-1}\frac{1}{4} = \tan^{-1}\frac{\frac{3}{5} + \frac{1}{4}}{1 - \frac{3}{5}.\frac{1}{4}}

    =tan⁡−117/2017/20=tan⁡−11=π4=\tan^{-1}\frac{17/20}{17/20} = \tan^{-1}1 = \frac{\pi}{4}

  29. We have to prove that tan⁡−12a−b3b+tan⁡−12b−a3a=π3\tan^{-1}\frac{2a - b}{\sqrt{3}b} + \tan^{-1}\frac{2b - a}{\sqrt{3}a} = \frac{\pi}{3}

    L.H.S. =tan⁡−12a−b3b+2b−a3a1−(2a−b)(2b−a)3ab= \tan^{-1}\frac{\frac{2a - b}{\sqrt{3}b} + \frac{2b - a}{\sqrt{3}a}}{1 - \frac{(2a - b)(2b - a)}{3ab}}

    =tan⁡−123a2−3ab+23b2−3ab3ab3ab−4ab+2a2+2b2−ab3ab= \tan^{-1}\frac{\frac{2\sqrt{3}a^2 - \sqrt{3}ab + 2\sqrt{3}b^2 -\sqrt{3}ab}{3ab}}{\frac{3ab - 4ab + 2a^2 + 2b^2 - ab}{3ab}}

    =tan⁡−12a2+23b2−23ab2a2+2b2−2ab=tan⁡−13=π3= \tan^{-1}\frac{2\sqrt{a^2} + 2\sqrt{3}b^2 - 2\sqrt{3}ab}{2a^2 + 2b^2 - 2ab} = \tan^{-1}\sqrt{3} = \frac{\pi}{3}

  30. We have to prove that tan⁡−125+tan⁡−113+tan⁡−1112=π4\tan^{-1}\frac{2}{5} + \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{12} = \frac{\pi}{4}

    L.H.S. =tan⁡−125+tan⁡−113+1121−13.112= \tan^{-1}\frac{2}{5} + \tan^{-1}\frac{\frac{1}{3} + \frac{1}{12}}{1 - \frac{1}{3}.\frac{1}{12}}

    =tan⁡−125+tan⁡−15/1235/36=tan⁡−125+tan⁡−137= \tan^{-1}\frac{2}{5} + \tan^{-1}\frac{5/12}{35/36} = \tan^{-1}\frac{2}{5} + \tan^{-1}\frac{3}{7}

    =tan⁡−125+371−25.37= \tan^{-1}\frac{\frac{2}{5} + \frac{3}{7}}{1 - \frac{2}{5}.\frac{3}{7}}

    =tan⁡−129/3529/35=tan⁡−11=π4= \tan^{-1}\frac{29/35}{29/35} = \tan^{-1}1 = \frac{\pi}{4}

  31. We have to prove that 2tan⁡−115+tan⁡−114=tan⁡−132432\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{4} = \tan^{-1}\frac{32}{43}

    L.H.S. =tan⁡−12.151−152+tan⁡−114= \tan^{-1}\frac{2.\frac{1}{5}}{1 - \frac{1}{5^2}} + \tan^{-1}\frac{1}{4}

    =tan⁡−1512+tan⁡−114= \tan^{-1}\frac{5}{12} + \tan^{-1}\frac{1}{4}

    =tan⁡−1512+141−512.14= \tan^{-1}\frac{\frac{5}{12} + \frac{1}{4}}{1 - \frac{5}{12}.\frac{1}{4}}

    =tan⁡−12/343/48=tan⁡−13243= \tan^{-1}\frac{2/3}{43/48} = \tan^{-1}\frac{32}{43}

  32. We have to prove that tan⁡−11+tan⁡−12+tan⁡−13=π=2(tan⁡−11+tan⁡−112+tan⁡−113)\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = \pi = 2\left(\tan^{-1}1 + \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}\right)

    tan⁡−11+tan⁡−12+tan⁡−13=tan⁡−11+tan⁡−12+31−2.3=tan⁡−11+tan⁡−1(−1)\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = \tan^{-1}1 + \tan^{-1}\frac{2 + 3}{1 - 2.3} = \tan^{-1}1 + \tan^{-1}(-1)

    =tan⁡−11−11+1=tan⁡−10=nπ= \tan^{-1}\frac{1 - 1}{1 + 1} =\tan^{-1}0 = n\pi

    2(tan⁡−11+tan⁡−112+tan⁡−113)2\left(\tan^{-1}1 + \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}\right)

    =2(tan⁡−11+tan⁡−112+121−12.13)= 2\left(\tan^{-1}1 + \tan^{-1}\frac{\frac{1}{2} + \frac{1}{2}}{1 - \frac{1}{2}.\frac{1}{3}}\right)

    =2(π4+tan⁡−11)=2.π2=π= 2\left(\frac{\pi}{4} + \tan^{-1}1\right) = 2.\frac{\pi}{2} = \pi

    Thus, the above expression will have principal value as pi.pi.

  33. We have to prove that tan⁡−1x+cot⁡−1y=tan⁡−1xy+1y−x\tan^{-1}x + \cot^{-1}y = \tan^{-1}\frac{xy + 1}{y - x}

    L.H.S. =tan⁡−1x+cot⁡−1y=tan⁡−1x+tan⁡−11y= \tan^{-1}x + \cot^{-1}y = \tan^{-1}x + \tan^{-1}\frac{1}{y}

    =tan⁡−1x+1y1−x.1y=tan⁡−1xy+1y−x= \tan^{-1}\frac{x + \frac{1}{y}}{1 - x.\frac{1}{y}} = \tan^{-1}\frac{xy + 1}{y - x}

  34. We have to prove that tan⁡−11x+y+tan⁡−1yx2+xy+1=cot⁡−1x\tan^{-1}\frac{1}{x + y} + \tan^{-1}\frac{y}{x^2 + xy + 1} = \cot^{-1}x

    L.H.S. =tan⁡−11x+y+tan⁡−1yx2+xy+1= \tan^{-1}\frac{1}{x + y} + \tan^{-1}\frac{y}{x^2 + xy + 1}

    =tan⁡−11x+y+yx2+xy+11−1x+y.yx2+xy+1= \tan^{-1}\frac{\frac{1}{x + y} + \frac{y}{x^2 + xy + 1}}{1 - \frac{1}{x + y}.\frac{y}{x^2 + xy + 1}}

    =tan⁡−1x2+2xy+y2+1x3+2x2y+xy2+x=tan⁡−11x=cot⁡−1x= \tan^{-1}\frac{x^2 + 2xy + y^2 + 1}{x^3 + 2x^2y + xy^2 + x} = \tan^{-1}\frac{1}{x} = \cot^{-1}x

  35. We have to prove that 2cot⁡−15+cot⁡−17+2cot⁡−18=π/42\cot^{-1}5 + \cot^{-1}7 + 2\cot^{-1}8 = \pi/4

    We know that cot⁡−1x+cot⁡−1y=xy−1x+y\cot^{-1}x + \cot^{-1}y = \frac{xy - 1}{x + y}

    ∴2(cot⁡−15+cot⁡−18)=2cot⁡−13913=2cos⁡−13=cot⁡−143\therefore 2\left(\cot^{-1}5 + \cot^{-1}8\right) = 2\cot^{-1}\frac{39}{13} = 2\cos^{-1}3 = \cot^{-1}\frac{4}{3}

    ∴2cot⁡−15+cot⁡−17+2cot⁡−18=cot⁡−143+cot⁡−17\therefore 2\cot^{-1}5 + \cot^{-1}7 + 2\cot^{-1}8 = \cot^{-1}\frac{4}{3} + \cot^{-1}7

    =cot⁡−1283−1253=cot⁡−11=π/4=\cot^{-1}\frac{\frac{28}{3} - 1}{\frac{25}{3}} = \cot^{-1}1 = \pi/4

  36. We have to prove that tan⁡−1a−b1+ab+tan⁡−1b−c1+bc+tan⁡−1c−a1+ca=0\tan^{-1}\frac{a - b}{1 + ab} + \tan^{-1}\frac{b - c}{1 + bc} + \tan^{-1}\frac{c - a}{1 + ca} = 0

    L.H.S. =tan⁡−1a−tan⁡−1b+tan⁡−1b−tan⁡−1c+tan⁡−1c−tan⁡−1a=0= \tan^{-1}a - \tan^{-1}b + \tan^{-1}b - \tan^{-1}c + \tan^{-1}c - \tan^{-1}a = 0

  37. We have to prove that tan⁡−1a3−b31+a3b3+tan⁡−1b3−c31+b3c3+tan⁡−1c3−a31+c3a3=0\tan^{-1}\frac{a^3 - b^3}{1 + a^3b^3} + \tan^{-1}\frac{b^3 - c^3}{1 + b^3c^3} + \tan^{-1}\frac{c^3 - a^3}{1 + c^3a^3} = 0

    L.H.S. =tan⁡−1a3−tan⁡−1b3+tan⁡−1b3−tan⁡−1c3+tan⁡−1c3−tan⁡−1a3=0= \tan^{-1}a^3 - \tan^{-1}b^3 + \tan^{-1}b^3 - \tan^{-1}c^3 + \tan^{-1}c^3 - \tan^{-1}a^3 = 0

  38. We have to prove that cot⁡−1xy+1y−x+cot⁡−1yz+1z−y+cot⁡−1z=tan⁡−11x\cot^{-1}\frac{xy + 1}{y - x} + \cot^{-1}\frac{yz + 1}{z - y} + \cot^{-1}z = \tan^{-1}\frac{1}{x}

    L.H.S. =cot⁡−1x−cot⁡−1y+cot⁡−1y−cot⁡−1z+cot⁡−1z=cot⁡−1x=tan⁡−11x= \cot^{-1}x - \cot^{-1}y + \cot^{-1}y - \cot^{-1}z + \cot^{-1}z= \cot^{-1}x = \tan^{-1}\frac{1}{x}

  39. We have to prove that cos⁡−1(cos⁡θ+cos⁡ϕ1+cos⁡θcos⁡ϕ)=2tan⁡−1(tan⁡θ2tan⁡ϕ2)\cos^{-1}\left(\frac{\cos\theta + \cos\phi}{1 + \cos\theta\cos\phi}\right) = 2\tan^{-1}\left(\tan\frac{\theta}{2}\tan\frac{\phi}{2}\right)

    L.H.S. =cos⁡−1(cos⁡θ+cos⁡ϕ1+cos⁡θcos⁡ϕ)= \cos^{-1}\left(\frac{\cos\theta + \cos\phi}{1 + \cos\theta\cos\phi}\right)

    =tan⁡−11+cos⁡2θcos⁡2ϕ+2cos⁡θcos⁡ϕ−cos⁡2θcos⁡2ϕ−2cos⁡θcos⁡ϕcos⁡θ+cos⁡ϕ= \tan^{-1}\frac{\sqrt{1 + \cos^2\theta\cos^2\phi + 2\cos\theta\cos\phi - \cos^2\theta\cos^2\phi - 2\cos\theta\cos\phi}}{\cos\theta + \cos\phi}

    =tan⁡−1(1−cos⁡2θ)(1−cos⁡2ϕ)cos⁡θ+cos⁡ϕ=tan⁡−1sin⁡θsin⁡ϕcos⁡θ+cos⁡ϕ= \tan^{-1}\frac{\sqrt{(1 - \cos^2\theta)(1 - \cos^2\phi)}}{\cos\theta + \cos\phi} = \tan^{-1}\frac{\sin\theta\sin\phi}{\cos\theta + \cos\phi}

    R.H.S. =2tan⁡−1(tan⁡θ2tan⁡ϕ2)= 2\tan^{-1}\left(\tan\frac{\theta}{2}\tan\frac{\phi}{2}\right)

    =tan⁡−12tan⁡θ2tan⁡ϕ21−tan⁡2θ2tan⁡2ϕ2= \tan^{-1}\frac{2\tan\frac{\theta}{2}\tan\frac{\phi}{2}}{1 - \tan^2\frac{\theta}{2}\tan^2\frac{\phi}{2}}

    =tan⁡−12tan⁡θ2tan⁡ϕ2.cos⁡2θ2cos⁡2ϕ2cos⁡2θ2cos⁡2ϕ2−sin⁡2θ2sin⁡2ϕ2=\tan^{-1}\frac{2\tan\frac{\theta}{2}\tan\frac{\phi}{2}.\cos^2\frac{\theta}{2}\cos^2\frac{\phi}{2}}{\cos^2\frac{\theta}{2}\cos^2\frac{\phi}{2} - \sin^2\frac{\theta}{2}\sin^2\frac{\phi}{2}}

    =tan⁡−112.sin⁡θsin⁡ϕcos⁡2θ2cos⁡2ϕ2−(1−cos⁡2θ2)(1−cos⁡2ϕ2)= \tan^{-1}\frac{1}{2}.\frac{\sin\theta\sin\phi}{\cos^2\frac{\theta}{2}\cos^2\frac{\phi}{2} - \left(1 - \cos^2\frac{\theta}{2}\right)\left(1-\cos^2\frac{\phi}{2}\right)}

    =tan⁡−1sin⁡θsin⁡ϕcos⁡θ+cos⁡ϕ= \tan^{-1}\frac{\sin\theta\sin\phi}{\cos\theta + \cos\phi}

  40. We have to prove that sin⁡−135+sin⁡−1817=sin⁡−17785\sin^{-1}\frac{3}{5} + \sin^{-1}\frac{8}{17} = \sin^{-1}\frac{77}{85}

    L.H.S. =sin⁡−135+sin⁡−1817= \sin^{-1}\frac{3}{5} + \sin^{-1}\frac{8}{17}

    =sin⁡−1(351−82172+8171−3252)=\sin^{-1}\left(\frac{3}{5}\sqrt{1 - \frac{8^2}{17^2}} + \frac{8}{17}\sqrt{1 - \frac{3^2}{5^2}}\right)

    =sin⁡−1(35.1517+817.45)= \sin^{-1}\left(\frac{3}{5}.\frac{15}{17} + \frac{8}{17}.\frac{4}{5}\right)

    sin⁡−1(45+3285)=sin⁡−17785=\sin^{-1}\left(\frac{45 + 32}{85}\right) = \sin^{-1}\frac{77}{85} = R.H.S.

  41. We have to prove that cos⁡−135+cos⁡−11213+cos⁡−16365=π2\cos^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13} + \cos^{-1}\frac{63}{65} = \frac{\pi}{2}

    We know that cos⁡−1x+cos⁡−1y=xy−(1−x2)(1−y2)\cos^{-1}x + \cos^{-1}y = xy - \sqrt{(1 - x^2)(1 - y^2)}

    L.H.S. =cos⁡−135+cos⁡−11213+cos⁡−16365= \cos^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13} + \cos^{-1}\frac{63}{65}

    =cos⁡−1(35.1213−(1−3252)(1−122132))+cos⁡−16365= \cos^{-1}\left(\frac{3}{5}.\frac{12}{13} - \sqrt{\left(1 - \frac{3^2}{5^2}\right)\left(1 - \frac{12^2}{13^2}\right)}\right) + \cos^{-1}\frac{63}{65}

    =cos⁡−1(3665−45.513)+cos⁡−16365= \cos^{-1}\left(\frac{36}{65} - \frac{4}{5}.\frac{5}{13}\right) + \cos^{-1}\frac{63}{65}

    =cos⁡−1(3665−2065)+cos⁡−16365= \cos^{-1}\left(\frac{36}{65} - \frac{20}{65}\right) + \cos^{-1}\frac{63}{65}

    =cos⁡−11665+cos⁡−16365= \cos^{-1}\frac{16}{65} + \cos^{-1}\frac{63}{65}

    =cos⁡−1(1665.6364−(1−162652)(1−632652))= \cos^{-1}\left(\frac{16}{65}.\frac{63}{64} - \sqrt{\left(1 - \frac{16^2}{65^2}\right)\left(1 - \frac{63^2}{65^2}\right)}\right)

    =cos⁡−10=π2== \cos^{-1}0 = \frac{\pi}{2} = R.H.S.

  42. We have to prove that sin⁡−1x+sin⁡−1y=cos⁡−1(1−x21−y2−xy)\sin^{-1}x + \sin^{-1}y = \cos^{-1}\left(\sqrt{1 - x^2}\sqrt{1 - y^2} - xy\right)

    L.H.S. =sin⁡−1x+sin⁡−1y=cos⁡−11−x2+cos⁡−11−y2= \sin^{-1}x + \sin^{-1}y = \cos^{-1}\sqrt{1 - x^2} + \cos^{-1}\sqrt{1 - y^2}

    =cos⁡1−(1−x21−y2−[1−(1−x2)][1−(1−y2)])= \cos^{1-}(\sqrt{1 - x^2}\sqrt{1 - y^2} - \sqrt{[1 - (1 - x^2)][1 - (1 - y^2)]})

    =cos⁡−1(1−x21−y2−xy)== \cos^{-1}\left(\sqrt{1 - x^2}\sqrt{1 - y^2} - xy\right) = R.H.S.

  43. We have to prove that 4(sin⁡−1110+cos⁡−125)=π4\left(\sin^{-1}\frac{1}{\sqrt{10}} + \cos^{-1}\frac{2}{\sqrt{5}}\right) =\pi

    or sin⁡−1110+cos⁡−125=π/4\sin^{-1}\frac{1}{\sqrt{10}} + \cos^{-1}\frac{2}{\sqrt{5}} =\pi/4

    L.H.S. =sin⁡−1110+sin⁡−115= \sin^{-1}\frac{1}{\sqrt{10}} + \sin^{-1}\frac{1}{\sqrt{5}}

    =sin⁡−1(1101−15+151−110)= \sin^{-1}\left(\frac{1}{\sqrt{10}}\sqrt{1 - \frac{1}{5}} + \frac{1}{\sqrt{5}}\sqrt{1 - \frac{1}{10}}\right)

    =sin⁡−1(250+350)=sin⁡−112= \sin^{-1}\left(\frac{2}{\sqrt{50}} + \frac{3}{\sqrt{50}}\right) = \sin^{-1}\frac{1}{\sqrt{2}}

    =π4== \frac{\pi}{4} = R.H.S.

  44. We have to prove that cos⁡(2sin⁡−1x)=1−2x2\cos(2\sin^{-1}x) = 1 - 2x^2

    L.H.S. =cos⁡[sin⁡−1(2x1−x2)]=cos⁡[cos⁡−11−4x2(1−x2)]=cos⁡[cos⁡−1(1−2x2)]= \cos[\sin^{-1}(2x\sqrt{1 - x^2})] = \cos[\cos^{-1}\sqrt{1 - 4x^2(1 - x^2)}] = \cos[\cos^{-1}(1 - 2x^2)]

    =1−2x2== 1 - 2x^2 = R.H.S.

  45. We have to prove that 12cos⁡−1x=sin⁡−11−x2=cos⁡−11+x2=tan⁡−11−x21+x\frac{1}{2}\cos^{-1}x = \sin^{-1}\sqrt{\frac{1 - x}{2}} = \cos^{-1}\sqrt{\frac{1 + x}{2}} = \tan^{-1}\frac{\sqrt{1 - x^2}}{1 + x}

    or cos⁡−1x=2sin⁡−11−x2=2cos⁡−11+x2=2tan⁡−11−x21+x\cos^{-1}x = 2\sin^{-1}\sqrt{\frac{1 - x}{2}} = 2\cos^{-1}\sqrt{\frac{1 + x}{2}} = 2\tan^{-1}\frac{\sqrt{1 - x^2}}{1 + x}

    2sin⁡−11−x2=sin⁡−1[2.1−x2.1−1−x2]2\sin^{-1}\sqrt{\frac{1 - x}{2}} = \sin^{-1}\left[2.\sqrt{\frac{1 - x}{2}}.\sqrt{1 - \frac{1 - x}{2}}\right]

    =sin⁡−12.1−x2.1+x2=sin⁡−11−x2=cos⁡−1x= \sin^{-1}2.\sqrt{\frac{1 - x}{2}}.\sqrt{\frac{1 + x}{2}} = \sin^{-1}\sqrt{1 - x^2} = \cos^{-1}x

    2cos⁡−11+x2=cos⁡−1[2.1+x2−1][∵2cos⁡−1x=cos⁡−1(2x2−1)]2\cos^{-1}\sqrt{\frac{1 + x}{2}} = \cos^{-1}\left[2.\frac{1 + x}{2} - 1\right][\because 2\cos^{-1}x = \cos^{-1}(2x^2 - 1)]

    =cos⁡−1x= \cos^{-1}x

    2tan⁡−11−x21+x=tan⁡−12.1−x2(1+x)1−1−x2(1+x)22\tan^{-1}\frac{\sqrt{1 - x^2}}{1 + x} = \tan^{-1}\frac{2.\frac{\sqrt{1 - x^2}}{(1 + x)}}{1 - \frac{1 - x^2}{(1 + x)^2}}

    =tan⁡−11−x2x=cos⁡−1x= \tan^{-1}\frac{\sqrt{1 - x^2}}{x} = \cos^{-1}x

  46. We have to prove that sin⁡−1x+cos⁡−1y=tan⁡−1xy+(1−x2)(1−y2)y1−x2−x1−y2\sin^{-1}x + \cos^{-1}y = \tan^{-1}\frac{xy + \sqrt{(1 - x^2)(1 - y^2)}}{y\sqrt{1 - x^2} - x\sqrt{1 - y^2}}

    L.H.S. =sin⁡−1x+cos⁡−1y=sin⁡−1x+sin⁡−11−y2= \sin^{-1}x + \cos^{-1}y = \sin^{-1}x + \sin^{-1}\sqrt{1 - y^2}

    =sin⁡−1[x1−(1−y2)+1−y21−x2]= \sin^{-1}[x\sqrt{1 -(1 - y^2)} + \sqrt{1 - y^2}\sqrt{1 - x^2}]

    =tan⁡−1xy+1−x21−y21−(xy+1−x2(1−y2))2=\tan^{-1}\frac{xy + \sqrt{1 - x^2}\sqrt{1 - y^2}}{\sqrt{1 - (xy + \sqrt{1 - x^2}(1 - y^2))^2}}

    =tan⁡−1xy+1−x21−y21−x2y2−(1−x2)(1−y2)−2xy1−x21−y2=\tan^{-1}\frac{xy + \sqrt{1 - x^2}\sqrt{1 - y^2}}{\sqrt{1 - x^2y^2 - (1 - x^2)(1 - y^2) - 2xy\sqrt{1 - x^2}\sqrt{1 - y^2}}}

    =tan⁡−1xy+1−x21−y2x2+y2−2xy1−x21−y62= \tan^{-1}\frac{xy + \sqrt{1 - x^2}\sqrt{1 - y^2}}{\sqrt{x^2 + y^2 - 2xy\sqrt{1 - x^2}\sqrt{1 - y62}}}

    =tan⁡−1xy+(1−x2)(1−y2)y1−x2−x1−y2= \tan^{-1}\frac{xy + \sqrt{(1 - x^2)(1 - y^2)}}{y\sqrt{1 - x^2} - x\sqrt{1 - y^2}}

  47. We have to prove that tan⁡−1x+tan⁡−1y=12sin⁡−12(x+y)(1−xy)(1+x2)(1+y2)\tan^{-1}x + \tan^{-1}y = \frac{1}{2}\sin^{-1}\frac{2(x + y)(1 - xy)}{(1 + x^2)(1 + y^2)}

    or 2(tan⁡−1x+tan⁡−1y=sin⁡−12(x+y)(1−xy)(1+x2)(1+y2)2(\tan^{-1}x + \tan^{-1}y = \sin^{-1}\frac{2(x + y)(1 - xy)}{(1 + x^2)(1 + y^2)}

    L.H.S. 2tan⁡−1x+y1−xy=tan⁡−12.x+y1−xy1−(x+y)2(1−xy)22\tan^{-1}\frac{x + y}{1 - xy} = \tan^{-1}\frac{2.\frac{x + y}{1 - xy}}{1 - \frac{(x + y)^2}{(1 - xy)^2}}

    =tan⁡−12(x+y)(1−xy)(1+x2y2−2xy−x2−Y2−2xy)= \tan^{-1}\frac{2(x + y)(1 - xy)}{(1 + x^2y^2 - 2xy - x^2 - Y^2 - 2xy)}

    =sin⁡−12(x+y)(1−xy)4(x+y)2(1−xy)2+(1+2x2y2−4xy−x2−y2)2= \sin^{-1}\frac{2(x + y)(1 - xy)}{\sqrt{4(x + y)^2(1 - xy)^2 + (1 + 2x^2y^2 - 4xy - x^2 - y^2)^2}}

    =sin⁡−12(x+y)(1−xy)(1+x2)(1+y2)= \sin^{-1}\frac{2(x + y)(1 - xy)}{(1 + x^2)(1 + y^2)}

  48. We have to prove that 2tan⁡−1(cosec⁡tan⁡−1x−tan⁡cot⁡−1x)=tan⁡−1x2\tan^{-1}(\cosec\tan^{-1}x - \tan\cot^{-1}x) = \tan^{-1}x

    L.H.S. =2tan⁡−1(cosec⁡tan⁡−1x−tan⁡cot⁡−1x)= 2\tan^{-1}(\cosec\tan^{-1}x - \tan\cot^{-1}x)

    =2tan⁡−1(cosec⁡cosec⁡−11+x2x−tan⁡tan⁡−11x)= 2\tan^{-1}\left(\cosec\cosec^{-1}\frac{\sqrt{1 + x^2}}{x} - \tan\tan^{-1}\frac{1}{x}\right)

    =2tan⁡−1(1+x2x−1x)=2tan⁡−11+x2−1x= 2\tan^{-1}\left(\frac{\sqrt{1 + x^2}}{x} - \frac{1}{x}\right) = 2\tan^{-1}\frac{\sqrt{1 + x^2} - 1}{x}

    =tan⁡−12.1+x2−1x1−(1+x2−1x)2= \tan^{-1}\frac{2.\frac{\sqrt{1 + x^2} - 1}{x}}{1 - \left(\frac{\sqrt{1 + x^2} - 1}{x}\right)^2}

    =tan⁡−1x= \tan^{-1}x

  49. We have to prove that cos⁡tan⁡−1sin⁡cot⁡−1x=x2+1x2+2\cos\tan^{-1}\sin\cot^{-1}x = \sqrt{\frac{x^2 + 1}{x^2 + 2}}

    L.H.S. =cos⁡tan⁡−1sin⁡cot⁡−1x=cos⁡tan⁡−1sin⁡sin⁡−111+x2= \cos\tan^{-1}\sin\cot^{-1}x = \cos\tan^{-1}\sin\sin^{-1}\frac{1}{\sqrt{1 + x^2}}

    =cos⁡tan⁡−111+x2=cos⁡cos⁡−11+x2x2+2= \cos\tan^{-1}\frac{1}{\sqrt{1 + x^2}} = \cos\cos^{-1}\frac{\sqrt{1 + x^2}}{\sqrt{x^2 + 2}}

    =x2+1x2+2= \sqrt{\frac{x^2 + 1}{x^2 + 2}}

  50. Clearly in a triangle A+B+C=πA + B + C = \pi where A,B,CA, B, C are angled of the triangle.

    Thus, π−C=A+B\pi - C = A + B

    Given A+B=tan⁡−12+tan⁡−13=tan⁡−12+31−2.3=tan⁡−1−1=3π4A + B = \tan^{-1}2 + \tan^{-1}3 = \tan^{-1}\frac{2 + 3}{1 - 2.3} = \tan^{-1}-1 = \frac{3\pi}{4}

    C=π−3π/4=π/4C = \pi - 3\pi/4 = \pi/4

  51. Given cos⁡−1x+cos⁡−1y+cos⁡−1z=π\cos^{-1}x + \cos^{-1}y + \cos^{-1}z = \pi

    ⇒cos⁡−1x+cos⁡−1y=π−cos⁡−1z\Rightarrow \cos^{-1}x + \cos^{-1}y = \pi - \cos^{-1}z

    ⇒xy−1−x21−y2=cos⁡(π−cos⁡−1z)=−z\Rightarrow xy - \sqrt{1 - x^2}\sqrt{1 - y^2} = \cos(\pi - \cos^{-1}z) = -z

    ⇒xy+z=1−x21−y2\Rightarrow xy + z = \sqrt{1 - x^2}\sqrt{1 - y^2}

    Squaring, we get

    x2y2+z2+2xyz=1−x2−y2+x2y2x^2y^2 + z^2 + 2xyz = 1 - x^2 - y^2 + x^2y^2

    ⇒x2+y2+z2+2xyz=1\Rightarrow x^2 + y^2 + z^2 + 2xyz = 1

  52. Given cos⁡−1x2+cos⁡−1y3=θ\cos^{-1}\frac{x}{2}+ \cos^{-1}\frac{y}{3} = \theta

    ⇒cos⁡−1[xy6−(4−x2)(9−y2)6]=θ\Rightarrow \cos^{-1}\left[\frac{xy}{6} - \frac{\sqrt{(4 - x^2)(9 - y^2)}}{6}\right] = \theta

    ⇒xy−6cos⁡θ=(4−x2)(9−y2)\Rightarrow xy - 6\cos\theta = \sqrt{(4 - x^2)(9 - y^2)}

    Squaring, we get

    x2y2+36cos⁡2θ−12xycos⁡θ=36−9x2−4y2+x2y2x^2y^2 + 36\cos^2\theta - 12xy\cos\theta = 36 -9x^2 - 4y^2 + x^2y^2

    ⇒9x2−12xycos⁡θ+4y2=36sin⁡2θ\Rightarrow 9x^2 - 12xy\cos\theta + 4y^2 = 36\sin^2\theta

  53. Let xryz=a,yrzx=b\sqrt{\frac{xr}{yz}} = a, \sqrt{\frac{yr}{zx}} = b and zrxy=c\sqrt{\frac{zr}{xy}} = c

    Then, L.H.S. =tan⁡−1a+tan⁡−1b+tan⁡−1c=a+b+c−abc1−ab−bc−ca= \tan^{-1}a + \tan^{-1}b + \tan^{-1}c = \frac{a + b + c- abc}{1 - ab - bc - ca}

    Now, a+b+c−abc=xr+yr+zrxyz−rrxyza + b + c - abc = \frac{x\sqrt{r} + y\sqrt{r} + z\sqrt{r}}{\sqrt{xyz}} - \frac{r\sqrt{r}}{\sqrt{xyz}}

    =r(x+y+z)−rrxyz=0= \frac{\sqrt{r}(x + y + z) - r\sqrt{r}}{\sqrt{xyz}} = 0

    and, 1−ab−bc−ca=1−r[1x+1y+1z]≠0[∵1x+1y+1z≠1r]1 - ab - bc - ca = 1- r\left[\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right]\neq 0[\because \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\neq \frac{1}{r}]

    ⇒\Rightarrow L.H.S. =0=nπ= 0 = n\pi and hence principal value is π\pi because sum of three positive angles cannot be zero or negative.

  54. Given u=cot⁡−1cos⁡2θ−tan⁡−1cos⁡2θu = \cot^{-1}\sqrt{\cos2\theta} - \tan^{-1}\sqrt{\cos2\theta}

    sin⁡u=sin⁡[cot⁡−1cos⁡2θ−tan⁡−1cos⁡2θ]\sin u = \sin[\cot^{-1}\sqrt{\cos2\theta} - \tan^{-1}\sqrt{\cos2\theta}]

    =sin⁡[tan⁡−11cos⁡2θ−tan⁡−1cos⁡2θ]= \sin\left[\tan^{-1}\frac{1}{\sqrt{\cos2\theta}} - \tan^{-1}\sqrt{\cos2\theta}\right]

    =sin⁡[tan⁡−11cos⁡2θ−cos⁡2θ1+1]=\sin\left[\tan^{-1}\frac{\frac{1}{\sqrt{\cos2\theta}} - \sqrt{\cos2\theta}}{1 + 1}\right]

    =sin⁡[tan⁡−1122sin⁡2θcos⁡2θ]=\sin\left[\tan^{-1}\frac{1}{2}\frac{2\sin^2\theta}{\sqrt{\cos2\theta}}\right]

    =sin⁡[sin⁡−1sin⁡2θsin⁡4θ+cos⁡2θ]= \sin\left[\sin^{-1}\frac{\sin^2\theta}{\sqrt{\sin^4\theta + \cos2\theta}}\right]

    =sin⁡[sin⁡−1sin⁡2θ(1−sin⁡2θ)]=sin⁡sin⁡−1tan⁡2θ=tan⁡2θ= \sin\left[\sin^{-1}\frac{\sin^2\theta}{(1 - \sin^2\theta)}\right] = \sin\sin^{-1}\tan^2\theta = \tan^2\theta

  55. Given cos⁡−1x3+cos⁡−1x=π2\cos^{-1}x\sqrt{3} + \cos^{-1}x = \frac{\pi}{2}

    cos⁡−1x3=π2−cos⁡−1x⇒cos⁡cos⁡−1x3=cos⁡(π2−cos⁡−1x)\cos^{-1}x\sqrt{3} = \frac{\pi}{2} - \cos^{-1}x \Rightarrow \cos\cos^{-1}x\sqrt{3} = \cos\left(\frac{\pi}{2} - \cos^{-1}x\right)

    ⇒x3=sin⁡cos⁡−1x=sin⁡sin⁡−11−x2\Rightarrow x\sqrt{3} = \sin\cos^{-1}x = \sin\sin^{-1}\sqrt{1 - x^2}

    ⇒x3=1−x2\Rightarrow x\sqrt{3} = \sqrt{1 - x^2}

    ⇒3x2=1−x2⇒x=±12\Rightarrow 3x^2 = 1 - x^2 \Rightarrow x = \pm\frac{1}{2}

    Case I: When x=12,x = \frac{1}{2}, given equation becomes

    cos⁡−132+cos⁡−112=π6+π3=π2\cos^{-1}\frac{\sqrt{3}}{2} + \cos^{-1}\frac{1}{2} = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}

    Case II: When x=−12,x = -\frac{1}{2},

    cos⁡−1−32+cos⁡−1−12=π−cos⁡−132+π−cos⁡−112\cos^{-1}-\frac{\sqrt{3}}{2} + \cos^{-1}-\frac{1}{2} = \pi - \cos^{-1}\frac{\sqrt{3}}{2} + \pi - \cos^{-1}\frac{1}{2}

    =3π2≠π2= \frac{3\pi}{2}\neq \frac{\pi}{2}

    Thus, x=12x = \frac{1}{2} is the only solution.

  56. Given equation is sin⁡−1x+sin⁡−12x=π3\sin^{-1}x + \sin^{-1}2x = \frac{\pi}{3}

    ⇒sin⁡−1x+sin⁡−12x=sin⁡−132\Rightarrow \sin^{-1}x + \sin^{-1}2x = \sin^{-1}\frac{\sqrt{3}}{2}

    ⇒sin⁡−1x−sin⁡−132=−sin⁡−12x\Rightarrow \sin{-1}x - \sin^{-1}\frac{\sqrt{3}}{2} = -\sin^{-1}2x

    ⇒sin⁡−1[x2−321−x2]=−sin⁡−12x\Rightarrow \sin^{-1}\left[\frac{x}{2} - \frac{\sqrt{3}}{2}\sqrt{1 - x^2}\right] = -\sin^{-1}2x

    ⇒x−3(1−x2)=−4x⇒25x2=3(1−x2)⇒x=±327\Rightarrow x - \sqrt{3(1 - x^2)} = -4x \Rightarrow 25x^2 = 3(1 - x^2) \Rightarrow x = \pm\frac{\sqrt{3}}{2\sqrt{7}}

    Clearly, x=−327x = -\frac{\sqrt{3}}{2\sqrt{7}} as angles will become negative and won’t satisfy the equality.

  57. Given, tan⁡−1x+tan⁡−1y+tan⁡−1z=π2,\tan^{-1}x + \tan^{-1}y + \tan^{-1}z= \frac{\pi}{2}, we have to prove that xy+yz+zx=1xy + yz + zx = 1

    tan⁡−1x+tan⁡−1y+tan⁡−1z=π2\tan^{-1}x + \tan^{-1}y + \tan^{-1}z= \frac{\pi}{2}

    ⇒tan⁡−1x+y+z−xyz1−xy−yz−zx=π2\Rightarrow \tan^{-1}\frac{x + y + z - xyz}{1 - xy - yz - zx} = \frac{\pi}{2}

    ⇒x+y+z−xyz1−xy−yz−zx=∞\Rightarrow \frac{x + y + z - xyz}{1 - xy - yz - zx} = \infty

    ⇒1−xy−yz−zx=0\Rightarrow 1 - xy - yz - zx = 0

    ⇒xy+yz+zx=1\Rightarrow x y + yz + zx = 1

  58. Given tan⁡−1x+tan⁡−1y+tan⁡−1z=π,\tan^{-1}x + \tan^{-1}y + \tan^{-1}z= \pi, we have to prove that x+y+z=xyzx + y + z = xyz

    tan⁡−1x+tan⁡−1y+tan⁡−1z=π\tan^{-1}x + \tan^{-1}y + \tan^{-1}z= \pi

    ⇒tan⁡−1x+y+z−xyz1−xy−yz−zx=π\Rightarrow \tan^{-1}\frac{x + y + z - xyz}{1 - xy - yz - zx} = \pi

    ⇒x+y+z−xyz1−xy−yz−zx=tan⁡π=0\Rightarrow \frac{x + y + z - xyz}{1 - xy - yz - zx} = \tan \pi = 0

    ⇒x+y+z=xyz\Rightarrow x + y + z = xyz

  59. Given sin⁡−1x+sin⁡−1y=π2,\sin^{-1}x + \sin^{-1}y = \frac{\pi}{2}, we have to prove that x1−y2+y1−x2=1x\sqrt{1 - y^2} + y\sqrt{1 - x^2} = 1

    ⇒sin⁡−1(x1−y2+y1−x2)=π2\Rightarrow \sin^{-1}(x\sqrt{1 - y^2} + y\sqrt{1 - x^2}) = \frac{\pi}{2}

    ⇒x1−y2+y1−x2=1\Rightarrow x\sqrt{1 - y^2} + y\sqrt{1 - x^2} = 1

  60. Give sin⁡−1x+sin⁡−1y+sin⁡−1z=π,\sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \pi, we have to prove that x1−x2+y1−y2+z1−z2=2xyzx\sqrt{1 - x^2} + y\sqrt{1 - y^2} + z\sqrt{1 - z^2} = 2xyz

    Let sin⁡−1x=A,sin⁡−1y=B\sin^{-1}x = A, \sin^{-1}y = B and sin⁡−1z=C\sin^{-1}z = C

    Then, A+B+C=π⇒A+B=π−CA + B + C = \pi \Rightarrow A + B = \pi - C

    We have to prove that sin⁡A1−sin⁡2A+B1−sin⁡2B+z1−sin⁡2C=2sin⁡Asin⁡Bsin⁡C\sin A\sqrt{1 - \sin^2A} + B\sqrt{1 - \sin^2B} + z\sqrt{1 - \sin^2C} = 2\sin A\sin B\sin C

    L.H.S. =sin⁡Acos⁡A+sin⁡Bcos⁡B+sin⁡Ccos⁡C= \sin A\cos A + \sin B\cos B + \sin C\cos C

    =12(sin⁡2A+sin⁡2B+sin⁡2C)=sin⁡(A+B)cos⁡(A−B)+sin⁡Ccos⁡[π−(A+B)]= \frac{1}{2}(\sin 2A + \sin 2B + \sin 2C) = \sin(A + B)\cos(A - B) + \sin C\cos[\pi - (A + B)]

    =sin⁡C[cos⁡(A−B)−cos⁡(A+B)][∵sin⁡(A+B)=sin⁡(π−C)=sin⁡C]= \sin C[\cos(A - B) - \cos(A + B)][\because \sin(A + B) = \sin(\pi - C) = \sin C]

    =2sin⁡Asin⁡Bsin⁡C== 2\sin A\sin B \sin C = R.H.S.

  61. Form given conditions 2tan⁡−1y=tan⁡x+tan⁡−1z2\tan^{-1}y = \tan^{x} + \tan^{-1}z and 2y=x+z2y = x + z

    ⇒2y1−y2=x+z1−zx\Rightarrow \frac{2y}{1 - y^2} = \frac{x + z}{1 - zx}

    ⇒1−y2=1−zx⇒y2=zx\Rightarrow 1 - y^2 = 1 - zx \Rightarrow y^2 = zx

    i.e. A.M. = G.M which is true only if x=y=zx = y = z

  62. Given cot⁡−1x+sin⁡−115=π4\cot^{-1}x + \sin^{-1}\frac{1}{\sqrt{5}} = \frac{\pi}{4}

    ⇒cot⁡−1x+cot⁡−1(5)2−1=π4\Rightarrow \cot^{-1}x + \cot^{-1}\sqrt{(\sqrt{5})^2 - 1} = \frac{\pi}{4}

    ⇒cot⁡−1x+cot⁡−12=π4\Rightarrow \cot^{-1}x + \cot^{-1}2 = \frac{\pi}{4}

    ⇒2x−1x+2=cot⁡π4=1⇒x=3\Rightarrow \frac{2x - 1}{x + 2} = \cot\frac{\pi}{4} = 1 \Rightarrow x = 3

  63. We have to solve tan⁡−12x+tan⁡−13x=π4\tan^{-1}2x + \tan^{-1}3x = \frac{\pi}{4}

    ⇒tan⁡−12x+3x1−2x.3x=π4\Rightarrow \tan^{-1}\frac{2x + 3x}{1 - 2x.3x} = \frac{\pi}{4}

    ⇒5x1−6x2=tan⁡π4=1\Rightarrow \frac{5x}{1 - 6x^2} = \tan\frac{\pi}{4} =1

    ⇒6x2+5x−1=0⇒(6x−1)(x+1)==0\Rightarrow 6x^2 + 5x - 1 = 0 \Rightarrow (6x - 1)(x + 1) = = 0

    ⇒x=−1,16\Rightarrow x = -1, \frac{1}{6}

    Clearly, x=−1x = -1 does not satisfy the equation ∴x=16\therefore x = \frac{1}{6}

  64. We have to solve tan⁡−1x+tan⁡−12x1−x2=π3\tan^{-1} x + \tan^{-1}\frac{2x}{1 - x^2} = \frac{\pi}{3}

    ⇒tan⁡−1x+2tan⁡−1x=π3\Rightarrow \tan^{-1}x + 2\tan^{-1}x = \frac{\pi}{3}

    ⇒3tan⁡−1x=π3\Rightarrow 3\tan^{-1}x = \frac{\pi}{3}

    ⇒x=tan⁡π9\Rightarrow x = \tan\frac{\pi}{9}

  65. We have to solve Solve tan⁡−112=cot⁡−1x+tan⁡−117\tan^{-1}\frac{1}{2} = \cot^{-1}x + \tan^{-1}\frac{1}{7}

    ⇒tan⁡−112−tan⁡−117=tan⁡−11x\Rightarrow \tan^{-1}\frac{1}{2} - \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{1}{x}

    ⇒tan⁡−112−171−12.17=tan⁡−11x\Rightarrow \tan^{-1}\frac{\frac{1}{2} - \frac{1}{7}}{1 - \frac{1}{2}.\frac{1}{7}} = \tan^{-1}\frac{1}{x}

    ⇒tan⁡−1513=tan⁡−11x\Rightarrow \tan^{-1}\frac{5}{13} = \tan^{-1}\frac{1}{x}

    ⇒x=135\Rightarrow x = \frac{13}{5}

  66. We have to solve tan⁡−1(x−1)+tan⁡−1x+tan⁡−1(x+1)=tan⁡−13x\tan^{-1}(x - 1) + \tan^{-1}x + \tan^{-1}(x + 1) = \tan^{-1}3x

    ⇒tan⁡−1(x−1)+tan⁡−1(x+1)=tan⁡−13x−tan⁡−1x\Rightarrow \tan^{-1}(x - 1) + \tan^{-1}(x + 1) = \tan^{-1}3x - \tan^{-1}x

    ⇒2x2−x2=2x1+3x2\Rightarrow \frac{2x}{2 - x^2} = \frac{2x}{1 + 3x^2}

    ⇒2x(4x2−1)=0\Rightarrow 2x(4x^2 - 1) = 0

    x=0,±12x = 0, \pm\frac{1}{2}

  67. We have to solve tan⁡−1x+1x−1+tan⁡−1x−1x=π+tan⁡−1(−7)\tan^{-1}\frac{x + 1}{x - 1} + \tan^{-1}\frac{x - 1}{x} = \pi + \tan^{-1}(-7)

    ⇒tan⁡−1x2+x+x2−2x+1x−x−1=π+tan⁡−1(−7)\Rightarrow \tan^{-1}\frac{x^2 + x + x^2 -2x + 1}{x - x - 1} = \pi + \tan^{-1}(-7)

    ⇒2x2−x+1=7x−7⇒2x2−8x+8=0\Rightarrow 2x^2 - x + 1 = 7x - 7 \Rightarrow 2x^2 - 8x + 8 = 0

    ⇒x2−4x+4=0⇒x=2\Rightarrow x^2 - 4x + 4 = 0 \Rightarrow x = 2

  68. We have to solve cot⁡−1(a−1)=cot⁡−1x+cot⁡−1(a2−x+1)\cot^{-1}(a - 1) = \cot^{-1}x + \cot^{-1}(a^2 - x + 1)

    cot⁡a−1cot⁡−1a2x−x2+x−1a2+1\cot^{a - 1} \cot^{-1}\frac{a^2x - x^2 + x - 1}{a^2 + 1}

    ⇒a3−a2+a−1=a2x−x2+x−1\Rightarrow a^3 - a^2 + a - 1 = a^2x - x^2 + x - 1

    ⇒x2−(1+a2)x+(a3−a2+a)=0\Rightarrow x^2 - (1 + a^2)x + (a^3 - a^2 + a) = 0

    ⇒(x−a)[x−(a2−a+1)]=0\Rightarrow (x - a)[x - (a^2 - a + 1)] = 0

    ⇒x=a,a2−a+1\Rightarrow x = a, a^2 - a + 1

  69. We have to solve sin⁡−12α1+α2+sin⁡−12β1+β2=2tan⁡−1x\sin^{-1}\frac{2\alpha}{1 + \alpha^2} + \sin^{-1}\frac{2\beta}{1 + \beta^2} = 2\tan^{-1}x

    We know that 2tan⁡−1x=sin⁡−12x1+x22\tan^{-1}x = \sin^{-1}\frac{2x}{1 + x^2}

    Thus, given equation becomes 2(tan⁡−1α+tan⁡−1β)=2tan⁡−1x2(\tan^{-1}\alpha + \tan^{-1}\beta) = 2\tan^{-1}x

    ⇒x=α+β1−αβ\Rightarrow x = \frac{\alpha + \beta}{1 - \alpha\beta}

  70. We have to solve cos⁡−1x2−1x2+1+tan⁡−12xx2−1=2π3\cos^{-1}\frac{x^2 - 1}{x^2 + 1} + \tan^{-1}\frac{2x}{x^2 - 1} = \frac{2\pi}{3}

    Case I: ⇒π−2tan⁡−1x−2tan⁡−1x=2π3\Rightarrow \pi - 2\tan^{-1}x - 2\tan^{-1}x = \frac{2\pi}{3}

    ⇒tan⁡−1x=π12\Rightarrow \tan^{-1}x = \frac{\pi}{12}

    x=2−3x = 2 - \sqrt{3}

    Case II: ⇒π−2tan⁡−1x+π−2tan⁡−1x=2π3\Rightarrow \pi - 2\tan^{-1}x + \pi - 2\tan^{-1}x = \frac{2\pi}{3}

    tan⁡−1x=π3⇒x=3\tan^{-1}x = \frac{\pi}{3} \Rightarrow x = \sqrt{3}

  71. We have to solve sin⁡−12a1+a2+cos⁡−11−b21+b2=2tan⁡−1x\sin^{-1}\frac{2a}{1 + a^2} + \cos^{-1}\frac{1 - b^2}{1 + b^2} = 2\tan^{-1}x

    ⇒2tan⁡−1x+2tan⁡−1b=2tan⁡−1x\Rightarrow 2\tan^{-1}x + 2\tan^{-1}b = 2\tan^{-1}x

    x=a+b1−abx = \frac{a + b}{1 - ab}

  72. We have to solve sin⁡−1x+sin⁡−1(1−x)=cos⁡−1x\sin^{-1}x + \sin^{-1}(1 - x) = \cos^{-1}x

    ⇒sin⁡−1(x2x−x2+(1−x)1−x2)=sin⁡−11−x2\Rightarrow \sin^{-1}\left(x\sqrt{2x - x^2} + (1 - x)\sqrt{1 - x^2}\right) = \sin^{-1}\sqrt{1 - x^2}

    ⇒x2x−x2+(1−x)1−x2=1−x2\Rightarrow x\sqrt{2x - x^2} + (1 - x)\sqrt{1 - x^2} = \sqrt{1 - x^2}

    ⇒x2x−x2=x1−x2\Rightarrow x\sqrt{2x - x^2} = x\sqrt{1 - x^2}

    Squaring, we get

    ⇒x2(2x−x2−1+x2)=0\Rightarrow x^2\left(2x - x^2 -1 + x^2\right) = 0

    x=0,12x = 0, \frac{1}{2}

  73. We have to solve tan⁡−1ax+12sec⁡−1bx=π4\tan^{-1}ax + \frac{1}{2}\sec^{-1}bx = \frac{\pi}{4}

    ⇒2tan⁡−1ax+sec−1bx=π2\Rightarrow 2\tan^{-1}ax + sec^{-1}bx = \frac{\pi}{2}

    ⇒tan⁡−12ax1−a2x2+tan⁡−11−b2x2=π2\Rightarrow \tan^{-1}\frac{2ax}{1 - a^2x^2} + \tan^{-1}\sqrt{1 - b^2x^2} = \frac{\pi}{2}

    ⇒tan⁡−12ax1−a2x2+1−b2x21−2ax1−a2x21−b2x2=π2\Rightarrow \tan^{-1}\frac{\frac{2ax}{1 - a^2x^2} + \sqrt{1 - b^2x^2}}{1 - \frac{2ax}{1 - a^2x^2}\sqrt{1 - b^2x^2}} = \frac{\pi}{2}

    ⇒1−2ax1−a2x21−b2x2=0\Rightarrow 1 - \frac{2ax}{1 - a^2x^2}\sqrt{1 - b^2x^2} = 0

    ⇒1−a2x2−2ax1−b2x2=0\Rightarrow 1 - a^2x^2 - 2ax\sqrt{1 - b^2x^2} = 0

    ⇒1−2a2x2+a4x4=4a2x2(1−b2x2)\Rightarrow 1 - 2a^2x^2 + a^4x^4 = 4a^2x^2(1 - b^2x^2)

    ⇒x=±12ab−a2\Rightarrow x = \pm \frac{1}{\sqrt{2ab - a^2}}

  74. We have to solve tan⁡(cos⁡−1x)=sin⁡(tan⁡−12)\tan(\cos^{-1}x) = \sin(\tan^{-1}2)

    ⇒tan⁡tan⁡−11−x2x=sin⁡sin⁡−124+1\Rightarrow \tan\tan^{-1}\frac{\sqrt{1 - x^2}}{x} = \sin\sin^{-1}\frac{2}{\sqrt{4 + 1}}

    ⇒1−x2x=25\Rightarrow \frac{\sqrt{1 - x^2}}{x} = \frac{2}{\sqrt{5}}

    ⇒5(1−x2)=4x2⇒x=±53\Rightarrow 5(1 - x^2) = 4x^2 \Rightarrow x = \pm\frac{\sqrt{5}}{3}

  75. We have to solve tan⁡(sec⁡−11x)=sin⁡cos⁡−115\tan\left(\sec^{-1}\frac{1}{x}\right) = \sin\cos^{-1}\frac{1}{\sqrt{5}}

    ⇒tan⁡tan⁡−11−x2x=sin⁡sin⁡−125\Rightarrow \tan\tan^{-1}\frac{\sqrt{1 - x^2}}{x} = \sin\sin^{-1}\frac{2}{\sqrt{5}}

    ⇒1−x2x2=45\Rightarrow \frac{1 - x^2}{x^2} = \frac{4}{5}

    ⇒x=±53\Rightarrow x = \pm\frac{\sqrt{5}}{3}

  76. We have to solve sin⁡−1x+sin⁡−1y=2π3\sin^{-1}x + \sin^{-1}y = \frac{2\pi}{3} and cos⁡−1x−cos⁡−1y=π3\cos^{-1}x - \cos^{-1}y = \frac{\pi}{3}

    sin⁡−1x+sin⁡−1y=2π3\sin^{-1}x + \sin^{-1}y = \frac{2\pi}{3}

    π2−cos⁡−1x+π2−cos⁡−1y=2π3\frac{\pi}{2} - \cos^{-1}x + \frac{\pi}{2} - \cos^{-1} y = \frac{2\pi}{3}

    ⇒cos⁡−1x+cos⁡−1y=π3\Rightarrow \cos^{-1}x + \cos^{-1}y = \frac{\pi}{3}

    Thus, 2cos⁡−1x=2π3⇒x=cos⁡π3=122\cos^{-1}x = 2\frac{\pi}{3} \Rightarrow x = \cos\frac{\pi}{3} = \frac{1}{2}

    and 2cos⁡−1y=0⇒y=12\cos^{-1}y = 0 \Rightarrow y = 1

  77. Let sin⁡−1(sin⁡10)=θ⇒sin⁡θ=sin⁡10=sin⁡35π11\sin^{-1}(\sin10) = \theta \Rightarrow \sin\theta = \sin10 = \sin\frac{35\pi}{11}

    sin⁡θ=sin⁡(3π+2π11)=−sin⁡2π11\sin\theta = \sin\left(3\pi + \frac{2\pi}{11}\right) = -\sin\frac{2\pi}{11}

    =sin⁡(−2π11)= \sin\left(-\frac{2\pi}{11}\right)

    θ=−2π11\theta = -\frac{2\pi}{11}

  78. 3tan⁡−1(12)=tan⁡−1[3.12−(12)31−3(12)2][∵tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ]3\tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left[\frac{3.\frac{1}{2} - \left(\frac{1}{2}\right)^3}{1 - 3\left(\frac{1}{2}\right)^2}\right]\left[\because \tan3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}\right]

    =tan⁡−1[11814]=tan⁡−1112= \tan^{-1}\left[\frac{\frac{11}{8}}{\frac{1}{4}}\right] = \tan^{-1}\frac{11}{2}

    2tan⁡−115=tan⁡−1[2.151−125]=tan⁡−15122\tan^{-1}\frac{1}{5} = \tan^{-1}\left[\frac{2.\frac{1}{5}}{1 - \frac{1}{25}}\right] = \tan^{-1}\frac{5}{12}

    Now 3tan⁡−112+2tan⁡−115=tan⁡−1112+tan⁡−15123\tan^{-1}\frac{1}{2} + 2\tan^{-1}\frac{1}{5} = \tan^{-1}\frac{11}{2} + \tan^{-1}\frac{5}{12}

    =π+tan⁡−1[112+5121−112.512]=π−tan⁡−114231= \pi + \tan^{-1}\left[\frac{\frac{11}{2} + \frac{5}{12}}{1- \frac{11}{2}.\frac{5}{12}}\right] = \pi - \tan^{-1}\frac{142}{31}

    Also, let sin⁡−1142655=θ\sin^{-1}\frac{142}{65\sqrt{5}} = \theta

    sin⁡θ=142655⇒tan⁡θ=14231\sin\theta = \frac{142}{65\sqrt{5}} \Rightarrow \tan\theta = \frac{142}{31}

    Thus, 3tan⁡−112+2tan⁡−115+sin⁡−1142655=π−tan⁡−114231+tan⁡−1142313\tan^{-1}\frac{1}{2} + 2\tan^{-1}\frac{1}{5} + \sin^{-1}\frac{142}{65\sqrt{5}} = \pi - \tan^{-1}\frac{142}{31} + \tan^{-1}\frac{142}{31}

    =π= \pi

  79. The given intervals indicate principal values of cos⁡−1x\cos^{-1}x and sin⁡−1x\sin^{-1}x.

    cos⁡[2cos⁡−1x+sin⁡−1x]=cos⁡(cos⁡−1x+cos⁡−1x+sin⁡−1x)\cos[2\cos^{-1}x + \sin^{-1}x] = \cos(\cos^{-1}x + \cos^{-1}x + \sin^{-1}x)

    =cos⁡[π2+cos⁡−1x]=−sin⁡cos⁡−1x=−sin⁡sin⁡−11−x2= \cos\left[\frac{\pi}{2} + \cos^{-1}x\right] = -\sin\cos^{-1}x = -\sin\sin^{-1}\sqrt{1 - x^2}

    =−1−x2=−1−125=−265= -\sqrt{1 - x^2} = -\sqrt{1 - \frac{1}{25}} = -\frac{2\sqrt{6}}{5}.

  80. We have to prove that 12cos⁡−135=tan⁡−112=π4−12cos⁡−145\frac{1}{2}\cos^{-1}\frac{3}{5} = \tan^{-1}\frac{1}{2} = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}\frac{4}{5}

    Let cos⁡−135=α,2tan⁡−112=β\cos^{-1}\frac{3}{5} = \alpha, 2\tan^{-1}\frac{1}{2} = \beta and π2−cos⁡−145=γ\frac{\pi}{2} - \cos^{-1}\frac{4}{5} = \gamma

    cos⁡α=cos⁡cos⁡−135=35\cos\alpha = \cos\cos^{-1}\frac{3}{5} = \frac{3}{5}

    cos⁡β=cos⁡[cos⁡−11−141+14]=cos⁡cos⁡−135=35\cos\beta = \cos\left[\cos^{-1}\frac{1 - \frac{1}{4}}{1 + \frac{1}{4}}\right] = \cos\cos^{-1}\frac{3}{5} = \frac{3}{5}

    cos⁡γ=cos⁡[π2−cos⁡−145]=sin⁡cos⁡−145=35\cos\gamma = \cos\left[\frac{\pi}{2} - \cos^{-1}\frac{4}{5}\right] = \sin\cos^{-1}\frac{4}{5} = \frac{3}{5}

    Thus, α=β=γ\alpha = \beta = \gamma

  81. Let A=2tan⁡−1(22−1)=2tan⁡−1(2×1.414−1)=2tan⁡−1(1.828)A = 2\tan^{-1}(2\sqrt{2} - 1) = 2\tan^{-1}(2\times 1.414 - 1) = 2\tan^{-1}(1.828)

    =2×(>60∘)[∵tan⁡60∘=3=1.732]= 2\times (> 60^\circ)[\because \tan60^\circ = \sqrt{3} = 1.732]

    Let B=3sin⁡−113+sin⁡−135B = 3\sin^{-1}\frac{1}{3} + \sin^{-1}\frac{3}{5}

    =sin⁡−1[3×13−4(13)3]+sin⁡−135= \sin^{-1}\left[3\times\frac{1}{3} - 4\left(\frac{1}{3}\right)^3\right] + \sin^{-1}\frac{3}{5}

    =sin⁡−12327+sin⁡−135=sin⁡−10.862+sin⁡−10.6= \sin^{-1}\frac{23}{27} + \sin^{-1}\frac{3}{5} = \sin^{-1}0.862 + \sin^{-1}0.6

    =<60∘+<45∘<105∘= <60^\circ + <45^\circ < 105^\circ

    Thus, AA is the greater angle.

  82. Whenever you have to sum trigonometric series of inverse terms check if it is possible to write them as difference of two terms and add the terms where terms cancel each other. If we look at the terms given in this series then that is possible.

    tan⁡−1(a1x−yx+a1y)=tan⁡−1(a1−yx1+a1yx)=tan⁡−1a1−tan⁡−1yx\tan^{-1}\left(\frac{a_1x - y}{x + a_1y}\right) = \tan^{-1}\left(\frac{a_1 - \frac{y}{x}}{1 + a_1\frac{y}{x}}\right) = \tan^{-1}a_1 - \tan^{-1}\frac{y}{x}

    tan⁡−1(a1−a11+a1a2)=tan⁡−1a2−tan⁡−1a1\tan^{-1}\left(\frac{a_1 - a_1}{1 + a_1a_2}\right) = \tan^{-1}a_2 - \tan^{-1}a_1

    …\ldots

    tan⁡−1(an−an−11+anan−1)=tan⁡−1an−tan⁡−1an−1\tan^{-1}\left(\frac{a_n - a_{n - 1}}{1 + a_na_{n - 1}}\right) = \tan^{-1}a_n - \tan^{-1}a_{n - 1}

    tan⁡−11an=cot⁡−1an\tan^{-1}\frac{1}{a_n} = \cot^{-1}a_n

    Adding these, we get L.H.S.=tan⁡−1an+cot⁡−1an−tan⁡−1yxL.H.S. = \tan^{-1}a_n + \cot^{-1}a_n - \tan^{-1}\frac{y}{x}

    =π2−tan⁡−1yx[∵tan⁡−1x+cot⁡−1x=π2]= \frac{\pi}{2} - \tan^{-1}\frac{y}{x}\left[\because\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}\right]

    =cot⁡−1yx=tan⁡−1xy=R.H.S.= \cot^{-1}\frac{y}{x} = \tan^{-1}\frac{x}{y} = R.H.S.

  83. Let tnt_n denote the nn-th term of the series, then tn=cot⁡−12n2=cot⁡−1(2n−1)−cot⁡−1(2n+1)t_n = \cot^{-1}2n^2 = \cot^{-1}(2n - 1) - \cot^{-1}(2n + 1)

    Putting n=1,2,3,..,n = 1,2,3, .., we get

    t1=cot⁡−11−cot⁡−13t_1 = \cot^{-1}1 - \cot^{-1}3

    t2=cot⁡−13−cot⁡−15t_2 = \cot^{-1}3 - \cot^{-1}5

    t3=cot⁡−15−cot⁡−17t_3 = \cot^{-1}5 - \cot^{-1}7

    …\ldots

    tn=cot⁡−1(2n−1)−cot⁡−1(2n+1)t_n = \cot^{-1}(2n - 1) - \cot^{-1}(2n + 1)

    Adding Sn=cot⁡−11−cot⁡−1(2n+1)S_n = \cot^{-1}1 - \cot^{-1}(2n + 1)

    As n→∞,cot⁡−1(2n+1)→0n\rightarrow \infty, \cot^{-1}(2n + 1)\rightarrow 0

    Hence, S∞=cot⁡−11=π4S_\infty = \cot^{-1}1 = \frac{\pi}{4}

  84. Case I. When x=1x = 1

    y=2tan⁡−1x+sin⁡−12x1+x2=2.tan⁡−11+sin⁡−121+1=2.π4+π2=πy = 2\tan^{-1}x + \sin^{-1}\frac{2x}{1 + x^2} = 2.\tan^{-1}1 + \sin^{-1}\frac{2}{1 + 1} = 2.\frac{\pi}{4} + \frac{\pi}{2} = \pi

    Case II. When x>1x > 1

    2tan⁡−1x=π−sin⁡−12x1+x2⇒y=π2\tan^{-1}x = \pi - \sin^{-1}\frac{2x}{1 + x^2} \Rightarrow y = \pi

  85. Let cos⁡−1x0=θ⇒cos⁡θ=x0\cos^{-1}x_0 = \theta \Rightarrow \cos\theta = x_0

    We are also given that xr+1=1+xr2x_{r + 1} = \sqrt{\frac{1 + x_r}{2}}

    Putting r=0,r = 0, we get x1=1+x02=1+cos⁡θ2x_1 = \sqrt{\frac{1 + x_0}{2}} = \sqrt{\frac{1 + \cos\theta}{2}}

    =cos⁡2θ2=∣cos⁡θ2∣=cos⁡θ2[∵0≤cos⁡−1x0≤π]= \sqrt{\cos^2\frac{\theta}{2}} = \left|\cos\frac{\theta}{2}\right| = \cos\frac{\theta}{2}[\because 0\leq\cos^{-1}x_0\leq \pi]

    Similarly, x2=1+cos⁡θ22=cos⁡θ22x_2 = \sqrt{\frac{1 + \cos\frac{\theta}{2}}{2}} = \cos\frac{\theta}{2^2}

    thus, xn=cos⁡θ2nx_n = \cos\frac{\theta}{2^n}

    Let y=x1x2x3…xny = x_1x_2x_3\ldots x_n then y=cos⁡θ2cos⁡θ22…cos⁡θ2ny = \cos\frac{\theta}{2}\cos\frac{\theta}{2^2}\ldots\cos\frac{\theta}{2^n}

    2ysin⁡θ2n=2sin⁡θ2ncos⁡θ2ncos⁡θ2n−1…cos⁡θ22y\sin\frac{\theta}{2^n} = 2\sin\frac{\theta}{2^n}\cos\frac{\theta}{2^n}\cos\frac{\theta}{2^{n - 1}}\ldots\cos\frac{\theta}{2}

    22yθ2n=2sin⁡θ2n−1cos⁡θ2n−1cos⁡θ2n−1…cos⁡θ22^2y\frac{\theta}{2^n} = 2\sin\frac{\theta}{2^{n - 1}}\cos\frac{\theta}{2^{n - 1}}\cos\frac{\theta}{2^{n - 1}}\ldots\cos\frac{\theta}{2}

    Proceeding like above, we finally arrive at following

    2n−1ysin⁡θ2n=sin⁡θ2cos⁡θ22^{n - 1}y\sin\frac{\theta}{2^n} = \sin\frac{\theta}{2}\cos\frac{\theta}{2}

    2nysin⁡θ2n=2sin⁡θ2cos⁡θ2=sin⁡θ2^ny\sin\frac{\theta}{2^n} = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} = \sin\theta

    y=12n.sin⁡θsin⁡θ2ny = \frac{1}{2^n}.\frac{\sin\theta}{\sin\frac{\theta}{2^n}}

    x1x2…x_1x_2\ldots to ∞=lim⁡n→∞12nsin⁡θsin⁡θ2n\infty = \lim_{n\to \infty} \frac{1}{2^n}\frac{\sin\theta}{\sin\frac{\theta}{2^n}}

    =lim⁡n→∞12nsin⁡θsin⁡θ2nθ2n.θ2n= \lim_{n\to\infty}\frac{1}{2^n}\frac{\sin\theta}{\frac{\sin\frac{\theta}{2^n}}{\frac{\theta}{2^n}}.\frac{\theta}{2^n}}

    =sin⁡θθ= \frac{\sin\theta}{\theta}

    R.H.S. =1−cos⁡2θsin⁡θθ=θ=cos⁡−1x0== \frac{\sqrt{1 - \cos^2\theta}}{\frac{\sin\theta}{\theta}} = \theta = \cos^{-1}x_0 = L.H.S.

  86. Let cos⁡−1ab=θ⇒cos⁡θ=an⇒a=bcos⁡θ\cos^{-1}\frac{a}{b} = \theta \Rightarrow \cos\theta = \frac{a}{n}\Rightarrow a = b\cos\theta

    Now, a1=a+b2=bcos⁡θ+b2=bcos⁡2θ2a_1 = \frac{a + b}{2} = \frac{b\cos\theta + b}{2} = b\cos^2\frac{\theta}{2}

    b1=a1b=bcos⁡2θ2.b=bcos⁡θ2b_1 = \sqrt{a_1b} = \sqrt{b\cos^2\frac{\theta}{2}.b} = b\cos\frac{\theta}{2}

    a2=a1+b12=bcos⁡2θ2+bcos⁡θ22=bcos⁡θ2cos⁡2θ22a_2 = \frac{a_1 + b_1}{2} = \frac{b\cos^2\frac{\theta}{2} + b\cos\frac{\theta}{2}}{2} = b\cos\frac{\theta}{2}\cos^2\frac{\theta}{2^2}

    b2=a2b1=bcos⁡θ2.cos⁡2θ22bcos⁡θ2=bcos⁡θ2cos⁡θ22b_2 = \sqrt{a_2b_1} = \sqrt{b\cos\frac{\theta}{2}.\cos^2\frac{\theta}{2^2}b\cos\frac{\theta}{2}} = b\cos\frac{\theta}{2}\cos\frac{\theta}{2^2}

    Proceeding as above, we get an=bcos⁡θ2cos⁡θ22…cos⁡θ2n=b.12n.sin⁡θsin⁡θ2na_n = b\cos\frac{\theta}{2}\cos\frac{\theta}{2^2}\ldots\cos\frac{\theta}{2^n} = b.\frac{1}{2^n}.\frac{\sin\theta}{\sin\frac{\theta}{2^n}}

    and bn=bcos⁡θ2cos⁡θ22…θ2nb_n = b\cos\frac{\theta}{2}\cos\frac{\theta}{2^2}\ldots\frac{\theta}{2^n}

    Now, lim⁡n→∞an=b.sin⁡θθ\lim_{n\to\infty}a_n = \frac{b.\sin\theta}{\theta} [like in previous problem]

    =b1−sin⁡2θcos⁡−1ab=b1−a2b2cos⁡−1ab=b2−a2cos⁡−1ab= \frac{b\sqrt{1 - \sin^2\theta}}{\cos^{-1}\frac{a}{b}} = \frac{b\sqrt{1 - \frac{a^2}{b^2}}}{\cos^{-1}\frac{a}{b}} = \frac{\sqrt{b^2 - a^2}}{\cos^{-1}\frac{a}{b}}

    and lim⁡n→∞bn=lim⁡n→∞bcos⁡θ2cos⁡θ22…cos⁡θ2n=b2−a2cos⁡−1ab\lim_{n\to\infty}b_n = \lim_{n\to\infty}b\cos\frac{\theta}{2}\cos\frac{\theta}{2^2}\ldots\cos\frac{\theta}{2^n} = \frac{\sqrt{b^2 - a^2}}{\cos^{-1}\frac{a}{b}}

  87. We have to prove that tan⁡−113+tan⁡−117+…+tan⁡−11n2+n+1=tan⁡−1nn+2\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + \ldots + \tan^{-1}\frac{1}{n^2 + n + 1} = \tan^{-1}\frac{n}{n + 2}

    When n=1n = 1, L.H.S. =tan⁡−113=\tan^{-1}\frac{1}{3} and R.H.S. =tan⁡−111+2=tan⁡−113= \tan^{-1}\frac{1}{1 + 2} = \tan^{-1}\frac{1}{3}

    We see that it is true for n=1n = 1. Let it is true for n=1n = 1

    ⇒tan⁡−113+tan⁡−117+…+tan⁡−11m2+m+1=tan⁡−1mm+2\Rightarrow \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + \ldots + \tan^{-1}\frac{1}{m^2 + m + 1} = \tan^{-1}\frac{m}{m + 2}

    Adding tan⁡−11(m+1)2+(m+1)+1\tan^{-1}\frac{1}{(m + 1)^2 + (m + 1) + 1} to both sides, we get

    R.H.S. =tan⁡−1mm+2+tan⁡−11(m+1)2+(m+1)+1= \tan^{-1}\frac{m}{m + 2} + \tan^{-1}\frac{1}{(m + 1)^2 + (m + 1) + 1}

    =tan⁡−1mm+1+tan⁡−1m+1m+3−tan⁡−1mm+2= \tan^{-1}\frac{m}{m + 1} + \tan^{-1}\frac{m + 1}{m + 3} - \tan^{-1}\frac{m}{m + 2}

    =tan⁡−1(m+1)+1(m+1)+2= \tan^{-1}\frac{(m + 1) + 1}{(m + 1) + 2}

    Thus, it is true for n=m+1n = m + 1 if it is true for n=mn = m. Hence, we have proven the result by using mathematical induction.

  88. Since x1,x2,x3,x4x_1, x_2, x_3, x_4 are the roots of the equation x4−x3sin⁡2β+x2cos⁡2β−xcos⁡β−sin⁡β=0x^4 - x^3\sin2\beta + x^2\cos2\beta - x\cos\beta - \sin\beta = 0

    ∴∑x1=x1+x2+x3+x4=−−sin⁡2β1=sin⁡2β\therefore \sum x_1 = x_1 + x_2 + x_3 + x_4 = -\frac{-\sin2\beta}{1} = \sin2\beta

    ∑x1x2=cos⁡2β\sum x_1x_2 = \cos2\beta

    ∑x1x2x3=cos⁡β\sum x_1x_2x_3 = \cos\beta

    and ∑x1x2x3x4=−sin⁡β\sum x_1x_2x_3x_4 = -\sin\beta

    Now tan⁡[tan⁡−1x1+tan⁡−1x2+tan⁡−1x3+tan⁡−1x4]=∑x1−∑x1x2x31−∑x1x2+x1x2x3x4\tan[\tan^{-1}x_1 + \tan^{-1}x_2 + \tan^{-1}x_3 + \tan^{-1}x_4] = \frac{\sum x_1 - \sum x_1x_2x_3}{1 - \sum x_1x_2 + x_1x_2x_3x_4}

    =sin⁡2β−cos⁡β1−cos⁡2β−sin⁡β=2sin⁡βcos⁡β−cos⁡β2sin⁡2β−sin⁡β= \frac{\sin2\beta - \cos\beta}{1 - \cos2\beta - \sin\beta} = \frac{2\sin\beta\cos\beta - \cos\beta}{2\sin^2\beta - \sin\beta}

    =cot⁡β= \cot\beta

    ⇒tan⁡[tan⁡−1x1+tan⁡−1x2+tan⁡−1x3+tan⁡−1x4]=tan⁡(π2−β)\Rightarrow \tan[\tan^{-1}x_1 + \tan^{-1}x_2 + \tan^{-1}x_3 + \tan^{-1}x_4] = \tan\left(\frac{\pi}{2} - \beta\right)

    ⇒tan⁡−1x1+tan⁡−1x2+tan⁡−1x3+tan⁡−1x4=nπ+π2−β\Rightarrow \tan^{-1}x_1 + \tan^{-1}x_2 + \tan^{-1}x_3 + \tan^{-1}x_4 = n\pi + \frac{\pi}{2} - \beta

  89. Let cot⁡−1(cot⁡5π4)=θ⇒cot⁡θ=cot⁡(π+π4)\cot^{-1}\left(\cot\frac{5\pi}{4}\right) = \theta \Rightarrow \cot\theta = \cot\left(\pi + \frac{\pi}{4}\right)

    =cot⁡π4⇒θ=π4= \cot\frac{\pi}{4}\Rightarrow \theta = \frac{\pi}{4}

  90. Let sin⁡−1(sin⁡5)=θ⇒sin⁡θ=sin⁡5=sin⁡35π22=sin⁡(π+13π22)\sin^{-1}(\sin5) = \theta \Rightarrow \sin\theta = \sin5 = \sin\frac{35\pi}{22} = \sin\left(\pi + \frac{13\pi}{22}\right)

    =−sin⁡13π22⇒sin⁡θ=−sin⁡13π22=−sin⁡(π−9π22)= -\sin\frac{13\pi}{22} \Rightarrow \sin\theta = -\sin\frac{13\pi}{22} = -\sin\left(\pi - \frac{9\pi}{22}\right)

    θ=−9π22=5−2π\theta = -\frac{9\pi}{22} = 5 - 2\pi

  91. Let cos⁡−1(cos⁡5π4)=θ⇒cos⁡θ=cos⁡(2π−3π4)\cos^{-1}(\cos\frac{5\pi}{4}) = \theta \Rightarrow \cos\theta = \cos\left(2\pi - \frac{3\pi}{4}\right)

    ⇒cos⁡θ=cos⁡3π4⇒θ=3π4\Rightarrow \cos\theta = \cos\frac{3\pi}{4} \Rightarrow \theta = \frac{3\pi}{4}

  92. Let cos⁡−1cos⁡10=θ⇒cos⁡θ=cos⁡10=cos⁡35π11=cos⁡(3π+2π11)\cos^{-1}\cos10 = \theta \Rightarrow \cos\theta = \cos10 = \cos\frac{35\pi}{11} = \cos\left(3\pi + \frac{2\pi}{11}\right)

    ⇒cos⁡θ=−cos⁡2π11=−cos⁡(π+−9π11)=cos⁡−9π11\Rightarrow \cos\theta = -\cos\frac{2\pi}{11} = -\cos\left(\pi + \frac{-9\pi}{11}\right) = \cos\frac{-9\pi}{11}

    ⇒θ=−9π11\Rightarrow \theta = \frac{-9\pi}{11}

  93. Given, sin⁡(2tan⁡−113)+cos⁡tan⁡−122\sin\left(2\tan^{-1}\frac{1}{3}\right) + \cos\tan^{-1}2\sqrt{2}

    =sin⁡tan⁡−12.131−19+cos⁡cos⁡−113= \sin\tan^{-1}\frac{2.\frac{1}{3}}{1 - \frac{1}{9}} + \cos\cos^{-1}\frac{1}{3}

    =sin⁡tan⁡−134+13=sin⁡sin⁡−135+13= \sin\tan^{-1}\frac{3}{4} + \frac{1}{3} = \sin\sin^{-1}\frac{3}{5} + \frac{1}{3}

    =35+13=1415= \frac{3}{5} + \frac{1}{3} = \frac{14}{15}

  94. Given, cot⁡[cot⁡−17+cot⁡−18+cot⁡−118]\cot[\cot^{-1}7 + \cot^{-1}8 + \cot^{-1}18]

    cot⁡−17+cot⁡−18+cot⁡−118=tan⁡−117+tan⁡−118+tan⁡−1118\cot^{-1}7 + \cot^{-1}8 + \cot^{-1}18 = \tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{8} + \tan^{-1}\frac{1}{18}

    =tan⁡−1(17+181−17.18)+tan⁡−1118=tan⁡−11555+tan⁡−1118= \tan^{-1}\left(\frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7}.\frac{1}{8}}\right) + \tan^{-1}\frac{1}{18}= \tan^{-1}\frac{15}{55} + \tan^{-1}\frac{1}{18}

    =tan⁡−1311+tan⁡−1118=tan⁡−1311+1181−311.118= \tan^{-1}\frac{3}{11} + \tan^{-1}\frac{1}{18} = \tan^{-1}\frac{\frac{3}{11} + \frac{1}{18}}{1 - \frac{3}{11}.\frac{1}{18}}

    =tan⁡−1.65198.198195=tan⁡−113=cot⁡−13= \tan^{-1}.\frac{65}{198}.\frac{198}{195} = \tan^{-1}\frac{1}{3} = \cot^{-1}3

    ∴cot⁡[cot⁡−17+cot⁡−18+cot⁡−118]=3\therefore \cot[\cot^{-1}7 + \cot^{-1}8 + \cot^{-1}18] = 3

  95. We have to prove that that sin⁡−135+cos⁡−11213+cot⁡−15633=π2\sin^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13} + \cot^{-1}\frac{56}{33} = \frac{\pi}{2}

    sin⁡−135=tan⁡−134\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{3}{4}

    cos⁡−11213=tan⁡−1512\cos^{-1}\frac{12}{13} = \tan^{-1}\frac{5}{12}

    cot⁡−15633=tan⁡−13356\cot^{-1}\frac{56}{33} = \tan^{-1}\frac{33}{56}

    ∴sin⁡−135+cos⁡−11213+cot⁡−15633=tan⁡−134+tan⁡−1512+tan⁡−13356\therefore \sin^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13} + \cot^{-1}\frac{56}{33} = \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{5}{12} + \tan^{-1}\frac{33}{56}

    =tan⁡−134+5121−34.512+tan⁡−13356= \tan^{-1}\frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4}.\frac{5}{12}} + \tan^{-1}\frac{33}{56}

    =tan⁡−15648.4833+tan⁡−13356=tan⁡−15633+tan⁡−13356= \tan^{-1}\frac{56}{48}.\frac{48}{33} + \tan^{-1}\frac{33}{56} = \tan^{-1}\frac{56}{33} + \tan^{-1}\frac{33}{56}

    We know that tan⁡−1x+tan⁡−11x=π/2∵\tan^{-1}x + \tan^{-1}\frac{1}{x} = \pi/2 \because denominator will be zero.

    Hence, tan⁡−15633+tan⁡−13356=π/2\tan^{-1}\frac{56}{33} + \tan^{-1}\frac{33}{56} = \pi/2

  96. We have to prove that 2cot⁡−15+cot⁡−17+2cot⁡−18=π42\cot^{-1}5 + \cot^{-1}7 + 2\cot^{-1}8 = \frac{\pi}{4}

    2cot⁡−16=2tan⁡−115=tan⁡−12.151−125=tan⁡−15122\cot^{-1}6 = 2\tan^{-1}\frac{1}{5} = \tan^{-1}\frac{2.\frac{1}{5}}{1 - \frac{1}{25}} = \tan^{-1}\frac{5}{12}

    2cot⁡−18=2tan⁡−118=tan⁡−12.181−164=tan⁡−116632\cot^{-1}8 = 2\tan^{-1}\frac{1}{8} = \tan^{-1}\frac{2.\frac{1}{8}}{1 - \frac{1}{64}} = \tan^{-1}\frac{16}{63}

    L.H.S. =tan⁡−1512+tan⁡−117+tan⁡−11663= \tan^{-1}\frac{5}{12} + \tan^{-1}\frac{1}{7} + \tan^{-1}\frac{16}{63}

    =tan⁡−1512+16631−512.1663+tan⁡−117=tan⁡−134+tan⁡−117= \tan^{-1}\frac{\frac{5}{12} + \frac{16}{63}}{1 - \frac{5}{12}.\frac{16}{63}} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{1}{7}

    =tan⁡−134+171−34.17=tan⁡−11=π4== \tan^{-1}\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4}.\frac{1}{7}} = \tan^{-1}1 = \frac{\pi}{4} = R.H.S.

  97. We have to prove that tan⁡−11+tan⁡−12+tan⁡−13=2(tan⁡−11+tan⁡−112+tan⁡−113).\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = 2\left(\tan^{-1}1 + \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}\right).

    L.H.S. =tan⁡−11+tan⁡−12+tan⁡−13=tan⁡−11+21−2+tan⁡−13= \tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = \tan^{-1}\frac{1 + 2}{1 - 2} + \tan^{-1}3

    =tan⁡−1(−3)+tan⁡−13=nπ= \tan^{-1}(-3) + \tan^{-1}3 = n\pi

    2tan⁡−11=tan⁡−11+11−1.1=tan⁡−1∞2\tan^{-1}1 = \tan^{-1}\frac{1 + 1}{1 - 1.1} = \tan^{-1}\infty

    2tan⁡−112=tan⁡−12.121−14=tan⁡−1432\tan^{-1}\frac{1}{2} = \tan^{-1}\frac{2.\frac{1}{2}}{1 - \frac{1}{4}} = \tan^{-1}\frac{4}{3}

    2tan⁡−113=tan⁡−12.131−19=tan⁡−1342\tan^{-1}\frac{1}{3} = \tan^{-1}\frac{2.\frac{1}{3}}{1 - \frac{1}{9}} = \tan^{-1}\frac{3}{4}

    Now tan⁡−1x+tan⁡−11x=2nπ+π2\tan^{-1}x + \tan^{-1}\frac{1}{x} = 2n\pi + \frac{\pi}{2}

    ∴\therefore R.H.S. =nπ= n\pi

  98. Given A=tan⁡−117A = \tan^{-1}\frac{1}{7} and B=tan⁡−113B = \tan^{-1}\frac{1}{3}, we have to prove that cos⁡2A=sin⁡4B\cos 2A = \sin 4B.

    cos⁡A=cos⁡tan⁡−117=cos⁡cos⁡−1750=750\cos A = \cos\tan^{-1}\frac{1}{7} = \cos\cos^{-1}\frac{7}{\sqrt{50}} = \frac{7}{\sqrt{50}}

    cos⁡2A=2cos⁡2A−1=2.4950−1=4850=2425\cos2A = 2\cos^2A - 1 = 2.\frac{49}{50} - 1 = \frac{48}{50} = \frac{24}{25}

    cos⁡B=cos⁡tan⁡−113=cos⁡cos⁡−1310⇒sin⁡B=110\cos B = \cos\tan^{-1}\frac{1}{3} = \cos\cos^{-1}\frac{3}{\sqrt{10}} \Rightarrow \sin B = \frac{1}{\sqrt{10}}

    sin⁡4B=4sin⁡Bcos⁡B(2cos⁡2B−1)=4.110.310(2.910−1)\sin4B = 4\sin B\cos B(2\cos^2B - 1) = 4.\frac{1}{\sqrt{10}}.\frac{3}{\sqrt{10}}\left(2.\frac{9}{10} - 1\right)

    =1210.810=2425= \frac{12}{10}.\frac{8}{10} = \frac{24}{25}

    Hence, cos⁡2A=sin⁡4B\cos2A = \sin4B.

  99. We have to find the sum tan⁡−1x1+1.2x2+tan⁡−1x1+2.3x2+…+tan⁡−111+n(n+1)x2,x>0.\tan^{-1}\frac{x}{1 + 1.2x^2} + \tan^{-1}\frac{x}{1 + 2.3x^2} + \ldots + \tan^{-1}\frac{1}{1 + n(n + 1)x^2}, x> 0.

    tan⁡−1x1+1.2x2=tan⁡−12x−tan⁡−1x\tan^{-1}\frac{x}{1 + 1.2x^2} = \tan^{-1}2x - \tan^{-1}x

    tan⁡−1x1+2.3x2=tan⁡−13x−tan⁡−12x\tan^{-1}\frac{x}{1 + 2.3x^2} = \tan^{-1}3x - \tan^{-1}2x

    …\ldots

    tan⁡−111+n(n+1)x2=tan⁡−1(n+1)x−tan⁡−1nx\tan^{-1}\frac{1}{1 + n(n + 1)x^2} = \tan^{-1}(n + 1)x - \tan^{-1}nx

    Adding, we get

    tan⁡−1x1+1.2x2+tan⁡−1x1+2.3x2+…+tan⁡−111+n(n+1)x2=tan⁡−1(n+1)x−tan⁡−1x\tan^{-1}\frac{x}{1 + 1.2x^2} + \tan^{-1}\frac{x}{1 + 2.3x^2} + \ldots + \tan^{-1}\frac{1}{1 + n(n + 1)x^2} = \tan^{-1}(n + 1)x - \tan^{-1}x

    =tan⁡−1nx1+(n+1)x2= \tan^{-1}\frac{nx}{1 + (n + 1)x^2}

  100. We have to find the sum tan⁡−1d1+a1a2+tan⁡−1d1+a2a3+…+tan⁡−1d1+anan+1,\tan^{-1}\frac{d}{1 + a_1a_2} + \tan^{-1}\frac{d}{1 + a_2a_3} + \ldots + \tan^{-1}\frac{d}{1 + a_na_{n + 1}}, where a1,a2,…,an,an+1a_1, a_2, \ldots, a_n, a_{n + 1} form an arithmetic progression with common difference d.d.

    tan⁡−1d1+a1a2=tan⁡−1a2−a11+a1a2=tan⁡−1a2−tan⁡−1a1\tan^{-1}\frac{d}{1 + a_1a_2} = \tan^{-1}\frac{a_2 - a_1}{1 + a_1a_2} = \tan^{-1}a_2 - \tan^{-1}a_1

    tan⁡−1d1+a2a3=tan⁡−1a3−a11+a2a3=tan⁡−1a3−tan⁡−1a2\tan^{-1}\frac{d}{1 + a_2a_3} = \tan^{-1}\frac{a_3 - a_1}{1 + a_2a_3} = \tan^{-1}a_3 - \tan^{-1}a_2

    …\ldots

    tan⁡−1d1+anan+1=tan⁡−1an+1−an1+anan+1=tan⁡−1an+1−an\tan^{-1}\frac{d}{1 + a_na_{n + 1}} = \tan^{-1}\frac{a_{n + 1} - a_n}{1 + a_na_{n +1}} = \tan^{-1}a_{n + 1} - a_n

    Adding, we get

    tan⁡−1d1+a1a2+tan⁡−1d1+a2a3+…+tan⁡−1d1+anan+1=tan⁡−1an+1−tan⁡−1a1=tan⁡−1nd1+a1an+1\tan^{-1}\frac{d}{1 + a_1a_2} + \tan^{-1}\frac{d}{1 + a_2a_3} + \ldots + \tan^{-1}\frac{d}{1 + a_na_{n + 1}} = \tan^{-1}a_{n + 1} - \tan^{-1}a_1 = \tan^{-1}\frac{nd}{1 + a_1a_{n + 1}}

  101. We have computed sin⁡−1sin⁡5=5−2π\sin^{-1}\sin5 = 5-2\pi, so we can rewrite the inequality as 5−2π>x2−4x5 - 2\pi>x^2 - 4x or x2−4x+2π−5<0x^2 - 4x + 2\pi - 5 < 0 which is a quadratic equation having positive coefficient for x2x^2. Thus it will be

    (x−α)(x−β)<0(x - \alpha)(x - \beta) < 0 for the above to hold true.

    ⇒[x−4−16−4(2π−5)2][x−4+16−4(2π−5)]<0\Rightarrow \left[x - \frac{4 - \sqrt{16 - 4(2\pi - 5)}}{2}\right]\left[x - \frac{4 + \sqrt{16 - 4(2\pi - 5)}}{}\right] < 0

    ⇒x∈(2−9−2π,2+9−2π)\Rightarrow x\in (2 - \sqrt{9 - 2\pi}, 2 + \sqrt{9 - 2\pi})

  102. Given, tan⁡−1y=5tan⁡−1x\tan^{-1}y = 5\tan^{-1}x, which we can rewrite as tan⁡−1y=2tan⁡−1x+3tan⁡−1x\tan^{-1}y = 2\tan^{-1}x + 3\tan^{-1}x

    R.H.S. =tan⁡−12x1−x2+tan⁡−13x−x31−3x2= \tan^{-1}\frac{2x}{1 - x^2} + \tan^{-1}\frac{3x - x^3}{1 - 3x^2}

    =tan⁡−12x(1−3x2)+(1−x2)(3x−x3)(1−x2)(1−3x2)−2x(3x−x3)= \tan^{-1}\frac{2x(1 - 3x^2) + (1 - x^2)(3x - x^3)}{(1 - x^2)(1 - 3x^2) - 2x(3x - x^3)}

    =tan⁡−12x−6x3+3x−x3−3x3+x51−4x2+3x4−6x2+2x4=\tan^{-1}\frac{2x - 6x^3 + 3x - x^3 -3x^3 + x^5}{1 - 4x^2 + 3x^4 - 6x^2 + 2x^4}

    ⇒y=x5−10x3+2x5x4−10x2+1\Rightarrow y = \frac{x^5 - 10x^3 + 2x}{5x^4 - 10x^2 + 1}

    Let tan⁡−1x=18∘\tan^{-1}x = 18^\circ then tan⁡−1y=π2⇒5x4−10x2+1=0\tan^{-1}y = \frac{\pi}{2}\Rightarrow 5x^4 - 10x^2 + 1 = 0.

  103. Let cos⁡−1x=α,cos⁡−1y=β,cos⁡−1z=γ\cos^{-1}x = \alpha, \cos^{-1}y = \beta, \cos^{-1}z = \gamma

    ⇒cos⁡α=x,cos⁡β=y,cos⁡γ=z\Rightarrow \cos\alpha = x, \cos\beta = y, \cos\gamma = z

    Also, given α+β+γ=π\alpha + \beta + \gamma = \pi

    and x+y+z=32⇒cos⁡α+cos⁡β+cos⁡γ=32x + y + z = \frac{3}{2} \Rightarrow \cos\alpha + \cos\beta + \cos\gamma = \frac{3}{2}

    Let z=cos⁡α+cos⁡β+cos⁡γz = \cos\alpha + \cos\beta + \cos\gamma and angle γ\gamma be fixed then

    z=2cos⁡α+β2cos⁡α−β2+cos⁡γz = 2\cos\frac{\alpha + \beta}{2}\cos\frac{\alpha - \beta}{2} + \cos\gamma

    =2sin⁡γ2cos⁡α−β2+cos⁡γ= 2\sin\frac{\gamma}{2}\cos\frac{\alpha - \beta}{2} + \cos\gamma

    Since γ\gamma is fixed, cos⁡γ\cos\gamma and sin⁡γ2\sin\frac{\gamma}{2} are fixed. Only changing term is cos⁡α−β2\cos\frac{\alpha - \beta}{2}

    Clearly, zz will be maximum if cos⁡α−β2=1\cos\frac{\alpha - \beta}{2} = 1 i.e. α=β\alpha = \beta

    Similarly, when angle β\beta is fixed, zz will be maximum if γ=α\gamma = \alpha

    and when angle α\alpha is fixed, zz will be maximum if β=γ\beta = \gamma

    ⇒z\Rightarrow z will be maximum if α+β+γ=60∘\alpha + \beta + \gamma = 60^\circ

    ⇒zmax=cos⁡60∘+cos⁡60∘+cos⁡60∘=32\Rightarrow z_{max} = \cos60^\circ + \cos60^\circ + \cos60^\circ = \frac{3}{2}

    ⇒α=β=γ=60∘⇒x=y=z\Rightarrow \alpha = \beta = \gamma = 60^\circ \Rightarrow x = y = z

  104. Let sin⁡−1x=α,sin⁡−1y=β,sin⁡−1z=γ\sin^{-1}x = \alpha, \sin^{-1}y = \beta, \sin^{-1}z = \gamma

    ⇒sin⁡α=x,sin⁡β=y,sin⁡γ=z\Rightarrow \sin\alpha = x, \sin\beta = y, \sin\gamma = z

    Also, α+β+γ=π\alpha + \beta + \gamma = \pi

    ⇒α+β=π−γ\Rightarrow \alpha + \beta = \pi - \gamma

    ⇒cos⁡(α+β)=cos⁡(π−γ)\Rightarrow \cos(\alpha + \beta) = \cos(\pi - \gamma)

    ⇒cos⁡αcos⁡β−sin⁡αsin⁡β=−cos⁡γ\Rightarrow \cos\alpha\cos\beta - \sin\alpha\sin\beta = -\cos\gamma

    ⇒1−x21−y2−xy=−1−z2\Rightarrow \sqrt{1 - x^2}\sqrt{1 - y^2} - xy = -\sqrt{1 - z^2}

    ⇒(1−x2)(1−y2)=xy−1−z2\Rightarrow \sqrt{(1 - x^2)(1 - y^2)} = xy - \sqrt{1 - z^2}

    Squaring, we get

    (1−x2)(1−y2)=x2y2+1−z2−2xy1−z2(1 - x^2)(1 - y^2) = x^2y^2 + 1 - z^2 -2xy\sqrt{1 - z^2}

    ⇒x2+y2−z2=2xy1−z2\Rightarrow x^2 + y^2 - z^2 = 2xy\sqrt{1 - z^2}

    Squaring again, we get

    x4+y4+z4+4x2y2z2=2(x2y2+y2z2+z2x2)x^4 + y^4 + z^4 + 4x^2y^2z^2 = 2(x^2y^2 + y^2z^2 + z^2x^2)

  105. Let tan⁡−1αβ=θ,tan⁡−1βα=ϕ\tan^{-1}\frac{\alpha}{\beta} = \theta, \tan^{-1}\frac{\beta}{\alpha} = \phi

    ∴tan⁡θ=αβ,tan⁡ϕ=βα\therefore \tan\theta = \frac{\alpha}{\beta}, \tan\phi = \frac{\beta}{\alpha}

    L.H.S. =α32sin⁡2θ2+β32cos⁡2ϕ2= \frac{\alpha^3}{2\sin^2\frac{\theta}{2}} + \frac{\beta^3}{2\cos^2\frac{\phi}{2}}

    =α31−cos⁡θ+β31+cos⁡ϕ= \frac{\alpha^3}{1 - \cos\theta} + \frac{\beta^3}{1 + \cos\phi}

    =α31−βα2+β2+β31+αα2+β2= \frac{\alpha^3}{1 - \frac{\beta}{\sqrt{\alpha^2 + \beta^2}}} + \frac{\beta^3}{1 + \frac{\alpha}{\sqrt{\alpha^2 + \beta^2}}}

    =α2+β2[α3(α2+β2+β)(α2+β2)−β2+β3(α2+β2)−α(α2+β2)−α2]= \sqrt{\alpha^2 + \beta^2}\left[\frac{\alpha^3(\sqrt{\alpha^2 + \beta^2}+ \beta)}{(\alpha^2 + \beta^2) - \beta^2} + \frac{\beta^3(\sqrt{\alpha^2 + \beta^2}) - \alpha}{(\alpha^2 + \beta^2) - \alpha^2}\right]

    =α2+β2[α(α2+β2+β)+β(α2+β2−α)]= \sqrt{\alpha^2 + \beta^2}[\alpha(\sqrt{\alpha^2 + \beta^2} + \beta) + \beta(\sqrt{\alpha^2 + \beta^2} - \alpha)]

    =(α2+β2)(α+β)= (\alpha^2 + \beta^2)(\alpha + \beta)

  106. We have to prove that 2tan⁡−1[tan⁡α2tan⁡(π4−β2)]=tan⁡−1[sin⁡αcos⁡βsin⁡β+cos⁡α].2\tan^{-1}\left[\tan\frac{\alpha}{2}\tan\left(\frac{\pi}{4} - \frac{\beta}{2}\right)\right] = \tan^{-1}\left[\frac{\sin\alpha\cos\beta}{\sin\beta + \cos\alpha}\right].

    L.H.S. =2tan⁡−1[tan⁡α2tan⁡(π4−β2)]= 2\tan^{-1}\left[\tan\frac{\alpha}{2}\tan\left(\frac{\pi}{4} - \frac{\beta}{2}\right)\right]

    =2tan⁡−1[tan⁡α21−tan⁡β21+tan⁡β2]= 2\tan^{-1}\left[\tan\frac{\alpha}{2}\frac{1 - \tan\frac{\beta}{2}}{1 + \tan\frac{\beta}{2}}\right]

    =tan⁡−1[2tan⁡α21−tan⁡β21+tan⁡β21−tan⁡2α2(1−tan⁡β2)2(1+tan⁡β2)2]= \tan^{-1}\left[\frac{2\tan\frac{\alpha}{2}\frac{1 - \tan\frac{\beta}{2}}{1 + \tan\frac{\beta}{2}}}{1 - \tan^2\frac{\alpha}{2}\frac{\left(1 - \tan\frac{\beta}{2}\right)^2}{\left(1 + \tan\frac{\beta}{2}\right)^2}}\right]

    Substituting tan⁡α2=sin⁡α2cos⁡α2\tan\frac{\alpha}{2} = \frac{\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}} and tan⁡β2=sin⁡α2cos⁡β2\tan\frac{\beta}{2} = \frac{\sin\frac{\alpha}{2}}{\cos\frac{\beta}{2}} and simplifying we arrive at the desired result.

  107. R.H.S. =tan⁡−1[tan⁡2(α+beta)tan⁡2(α−β)]+tan⁡−11= \tan^{-1}[\tan^2(\alpha + beta)\tan^2(\alpha - \beta)] + \tan^{-1}1

    =tan⁡−1[1+tan⁡2(α+β)tan⁡2(α−beta)1−tan⁡2(α+β)tan⁡2(α−beta)]= \tan^{-1}\left[\frac{1 + \tan^2(\alpha + \beta)\tan^{2}(\alpha - beta)}{1 - \tan^2(\alpha + \beta)\tan^{2}(\alpha - beta)}\right]

    =tan⁡−1[cos⁡2(α+β)cos⁡2(α−β)+sin⁡2(α+β)sin⁡2(α−β)cos⁡2(α+β)cos⁡2(α−β)−sin⁡2(α+β)sin⁡2(α−β)]= \tan^{-1}\left[\frac{\cos^2(\alpha + \beta)\cos^2(\alpha - \beta) + \sin^2(\alpha + \beta)\sin^2(\alpha - \beta)}{\cos^2(\alpha + \beta)\cos^2(\alpha - \beta) - \sin^2(\alpha + \beta)\sin^2(\alpha - \beta)}\right]

    =tan⁡−1[4cos⁡2(α+β)cos⁡2(α−β)+4sin⁡2(α+β)sin⁡2(α−β)4cos⁡2(α+β)cos⁡2(α−β)−4sin⁡2(α+β)sin⁡2(α−β)]= \tan^{-1}\left[\frac{4\cos^2(\alpha + \beta)\cos^2(\alpha - \beta) + 4\sin^2(\alpha + \beta)\sin^2(\alpha - \beta)}{4\cos^2(\alpha + \beta)\cos^2(\alpha - \beta) - 4\sin^2(\alpha + \beta)\sin^2(\alpha - \beta)}\right]

    =tan⁡−1[{2cos⁡(α+β)cos⁡(α−β)}2+{2sin⁡(α+β)sin⁡(α−β)}2{2cos⁡(α+β)cos⁡(α−β)}2−{2sin⁡(α+β)sin⁡(α−β)}2]= \tan^{-1}\left[\frac{\{2\cos(\alpha + \beta)\cos(\alpha - \beta)\}^2 + \{2\sin(\alpha + \beta)\sin(\alpha - \beta)\}^2}{\{2\cos(\alpha + \beta)\cos(\alpha - \beta)\}^2 - \{2\sin(\alpha + \beta)\sin(\alpha - \beta)\}^2}\right]

    =tan⁡−1[(cos⁡2α+cos⁡2β)2+(cos⁡2β−cos⁡2α)2cos⁡2α+cos⁡2β)2−(cos⁡2β−cos⁡2α)2]=\tan^{-1}\left[\frac{(\cos2\alpha + \cos2\beta)^2 + (\cos2\beta - \cos2\alpha)^2}{\cos2\alpha + \cos2\beta)^2 - (\cos2\beta - \cos2\alpha)^2}\right]

    =tan⁡−1[2cos⁡2α+2cos⁡2β4cos⁡2αcos⁡β]= \tan^{-1}\left[\frac{2\cos^2\alpha + 2\cos^2\beta}{4\cos2\alpha\cos\beta}\right]

    =tan⁡−1[12cos⁡2αsec⁡2β+12cos⁡βsec⁡α]== \tan^{-1}\left[\frac{1}{2}\cos2\alpha\sec2\beta + \frac{1}{2}\cos\beta\sec\alpha\right] = L.H.S.

  108. Let 3−4x2x2=t⇒3−4x24x2=t2\sqrt{\frac{3 - 4x^2}{x^2}} = t \Rightarrow \sqrt{\frac{3 - 4x^2}{4x^2}} = \frac{t}{2}

    ∴\therefore R.H.S. =2tan⁡−1t2−tan⁡−1t= 2\tan^{-1}\frac{t}{2} - \tan^{-1}t

    =tan⁡−1t1−t24−tan⁡−1t=tan⁡−14t4−t2−tan−1t= \tan{-1}\frac{t}{1 - \frac{t^2}{4}} - \tan^{-1}t = \tan^{-1}\frac{4t}{4 - t^2} - tan^{-1}t

    =tan⁡−14t4−t2−t1+4t24−t2=tan−1t34+3t2= \tan^{-1}\frac{\frac{4t}{4 - t^2} - t}{1 + \frac{4t^2}{4 - t^2}} = tan^{-1}\frac{t^3}{4 + 3t^2}

    ⇒cot⁡−1y1−x2−y2=tan⁡−1t34+3t2\Rightarrow \cot^{-1}\frac{y}{\sqrt{1 - x^2 - y^2}} = \tan^{-1}\frac{t^3}{4 + 3t^2}

    ⇒1−x2−y2y2=t69t4+24t2+16\Rightarrow \frac{1 - x^2 - y^2}{y^2} = \frac{t^6}{9t^4 + 24t^2 + 16}

    ⇒1−x2y2−1=t69t4+24t2+16\Rightarrow \frac{1 - x^2}{y^2} - 1 = \frac{t^6}{9t^4 + 24t^2 + 16}

    ⇒1−x2y2=t6+9t4+24t2+169t4+24t2+16\Rightarrow \frac{1 - x^2}{y^2} = \frac{t^6 + 9t^4 + 24t^2 + 16}{9t^4 + 24t^2 + 16}

    ⇒y2=9t4+24t2+16t6+9t4+24t2+16(1−x2)\Rightarrow y^2 = \frac{9t^4 + 24t^2 + 16}{t^6 + 9t^4 + 24t^2 + 16}(1 - x^2)

    We know that t2+4=3x2t^2 + 4 = \frac{3}{x^2} from our initial equation.

    ⇒y2=(t2+4)2+8t4+16t2(t2+1)(t2+4)2(1−x2)\Rightarrow y^2 = \frac{(t^2 + 4)^2 + 8t^4 + 16t^2}{(t^2 + 1)(t^2 + 4)^2}(1 - x^2)

    Substituting for tt and simplifying, we obtain

    27y2=81x2−144x4+64x627y^2 = 81x^2 -144x^4 + 64x^6

  109. Given mtan⁡(α−θ)cos⁡2θ=ntan⁡θcos⁡2(α−θ)\frac{m\tan(\alpha - \theta)}{\cos^2\theta} = \frac{n\tan\theta}{\cos^2(\alpha - \theta)}

    ⇒mn=sin⁡θcos⁡θsin⁡(α−θ)cos⁡(α−θ)=sin⁡2θsin⁡2(α−θ)\Rightarrow \frac{m}{n} = \frac{\sin\theta\cos\theta}{\sin(\alpha - \theta)\cos(\alpha - \theta)} = \frac{\sin2\theta}{\sin2(\alpha - \theta)}

    Doing componendo and dividendo, we get

    n−mm+n=sin⁡2(α−θ)−sin⁡2θsin⁡2θ+sin⁡2(α−θ)\frac{n - m}{m + n} = \frac{\sin2(\alpha - \theta) - \sin2\theta}{\sin2\theta + \sin2(\alpha - \theta)}

    ⇒n−mm+n=cos⁡αsin⁡(α−2θ)sin⁡αcos⁡(α−2θ)\Rightarrow \frac{n - m}{m + n} = \frac{\cos\alpha\sin(\alpha - 2\theta)}{\sin\alpha\cos(\alpha - 2\theta)}

    ⇒n−mm+n=tan⁡(α−2θ)tan⁡α\Rightarrow \frac{n - m}{m + n} = \frac{\tan(\alpha - 2\theta)}{\tan\alpha}

    ⇒α−2θ=tan⁡−1(n−mm+n)tan⁡α\Rightarrow \alpha - 2\theta = \tan^{-1}\left(\frac{n - m}{m + n}\right)\tan\alpha

    ⇒θ=12[α−tan⁡−1(n−mn+m)tan⁡α]\Rightarrow \theta = \frac{1}{2}\left[\alpha - \tan^{-1}\left(\frac{n - m}{n + m}\right)\tan\alpha\right]

  110. Given, sin⁡−1xa+sin⁡−1yb=sin⁡−1c2ab\sin^{-1}\frac{x}{a} + \sin^{-1}\frac{y}{b} = \sin^{-1}\frac{c^2}{ab}

    ⇒sin⁡−1xa=sin⁡−1c2ab−sin⁡−1yb\Rightarrow \sin^{-1}\frac{x}{a} = \sin^{-1}\frac{c^2}{ab} - \sin^{-1}\frac{y}{b}

    ⇒xa=c2ab1−y2b2−yb1−c4a2b2\Rightarrow \frac{x}{a} = \frac{c^2}{ab}\sqrt{1 - \frac{y^2}{b^2}} - \frac{y}{b}\sqrt{1 - \frac{c^4}{a^2b^2}}

    ⇒xa+yb1−c4a2b2=c2ab1−y2b2\Rightarrow \frac{x}{a} + \frac{y}{b}\sqrt{1 - \frac{c^4}{a^2b^2}} = \frac{c^2}{ab}\sqrt{1 - \frac{y^2}{b^2}}

    Squaring both sides

    ⇒x2a2+y2b2(1−c4a2b2)+2xyab1−c4a2b2=c4a2b2(1−y2b2)\Rightarrow \frac{x^2}{a^2} + \frac{y^2}{b^2}\left(1 - \frac{c^4}{a^2b^2}\right) + \frac{2xy}{ab}\sqrt{1 - \frac{c^4}{a^2b^2}} = \frac{c^4}{a^2b^2}\left(1 - \frac{y^2}{b^2}\right)

    ⇒b2x2+2xya2b2−c4=c4−a2y2\Rightarrow b^2x^2 + 2xy\sqrt{a^2b^2 - c^4} = c^4 - a^2y^2

  111. We have to prove that tan⁡−1t+tan⁡−12t1−t2=tan⁡−13t−t31−3t2\tan^{-1}t + \tan^{-1}\frac{2t}{1 - t^2} = \tan^{-1}\frac{3t - t^3}{1 - 3t^2}

    L.H.S. =tan⁡−1t+2tan⁡−1t=3tan⁡−1t=tan⁡−13t−t31−3t2= \tan^{-1}t + 2\tan^{-1}t = 3\tan^{-1}t = \tan^{-1}\frac{3t - t^3}{1 - 3t^2}

  112. If a>x>ba>x>b or a<x<ba<x<b then the fractions under square root are positive and less than one. So the angles are defined.

    cos⁡−1a−xa−b=sin⁡−11−a−xa−b=sin⁡−1x−ba−b\cos^{-1}\sqrt{\frac{a - x}{a - b}} = \sin^{-1}\sqrt{1 - \frac{a - x}{a - b}} = \sin^{-1}\sqrt{\frac{x - b}{a - b}}

  113. Given, cos⁡−1p+cos⁡−11−p+cos⁡−11−q=3π4\cos^{-1}\sqrt{p} + \cos^{-1}\sqrt{1 - p} + \cos^{-1}\sqrt{1 - q} = \frac{3\pi}{4}.

    ⇒cos⁡−1p+sin⁡−1p+cos⁡−11−q=3π4\Rightarrow \cos^{-1}\sqrt{p} + \sin^{-1}\sqrt{p} + \cos^{-1}\sqrt{1 - q} = \frac{3\pi}{4}

    ⇒π2+cos⁡−11−q=3π4\Rightarrow \frac{\pi}{2} + \cos^{-1}\sqrt{1 - q} = \frac{3\pi}{4}

    ⇒cos⁡−11−q=π4⇒1−q=12⇒q=12\Rightarrow \cos^{-1}\sqrt{1 - q} = \frac{\pi}{4}\Rightarrow 1 - q = \frac{1}{2} \Rightarrow q = \frac{1}{2}

    For cos⁡−1p\cos^{-1}\sqrt{p} to be defined 0≤p≤10\leq p\leq 1 and then cos⁡−11−p\cos^{-1}\sqrt{1 - p} will also be defined.

  114. Given, tan⁡−1x+cot⁡−1y=tan⁡−13\tan^{-1}x + \cot^{-1}y = \tan^{-1}3

    ⇒tan⁡−1x+tan⁡−11y=tan⁡−13\Rightarrow \tan^{-1}x + \tan^{-1}\frac{1}{y} = \tan^{-1}3

    ⇒tan⁡−1x+1y1−xy=tan⁡−13\Rightarrow \tan^{-1}\frac{x + \frac{1}{y}}{1 - \frac{x}{y}} = \tan^{-1}3

    ⇒xy+1y−x=3⇒y=3x+13−x\Rightarrow \frac{xy + 1}{y - x} = 3 \Rightarrow y = \frac{3x + 1}{3 - x}

    When xx is positive, numerator is positive. For denominator to be positive x=1,2x = 1, 2 (considering only integral values). Corresponding values of y=2,7y = 2, 7. We can see that both solutions satisfy the original equation.

  115. Given sin⁡−1axc+sin⁡−1bxc=sin⁡−1x\sin^{-1}\frac{ax}{c} + \sin^{-1}\frac{bx}{c} = \sin^{-1}x so we can infer −1≤x≤1-1\leq x\leq 1

    Also given, a2+b2=c2⇒a2x2c2+b2x2c2=x2a^2 + b^2 = c^2 \Rightarrow \frac{a^2x^2}{c^2} + \frac{b^2x^2}{c^2} = x^2

    From first equation, axc1−b2x2c2+bxc1−a2x2c2=x\frac{ax}{c}\sqrt{1 - \frac{b^2x^2}{c^2}} + \frac{bx}{c}\sqrt{1 - \frac{a^2x^2}{c^2}} = x

    ⇒x[axc1−b2x2c2+bxc1−a2x2c2−1]=0\Rightarrow x\left[\frac{ax}{c}\sqrt{1 - \frac{b^2x^2}{c^2}} + \frac{bx}{c}\sqrt{1 - \frac{a^2x^2}{c^2}} - 1\right] = 0

    Either x=0x = 0 or ac1−b2x2c2+bc1−a2x2c2=0\frac{a}{c}\sqrt{1 - \frac{b^2x^2}{c^2}} + \frac{b}{c}\sqrt{1 - \frac{a^2x^2}{c^2}} = 0

    ⇒ac2−b2x2+bc2−a2x2=c2\Rightarrow a\sqrt{c^2 - b^2x^2} + b\sqrt{c^2 - a^2x^2} = c^2

    ⇒ac2−b2x2=c2−bc2−a2x2\Rightarrow a\sqrt{c^2 - b^2x^2} = c^2 - b\sqrt{c^2 - a^2x^2}

    ⇒a2c2−a2b2x2=c4+b2c2−a2b2x2−2bc2c2−a2x2\Rightarrow a^2c^2 - a^2b^2x^2 = c^4 + b^2c^2 - a^2b^2x^2 - 2bc^2\sqrt{c^2 - a^2x^2}

    ⇒a2c2−b2c2−c4=2bc2c2−a2x2\Rightarrow a^2c^2 - b^2c^2 - c^4 = 2bc^2\sqrt{c^2 - a^2x^2}

    ⇒a2−b2−c2=−2bc2−a2x2\Rightarrow a^2 - b^2 - c^2 = -2b\sqrt{c^2 - a^2x^2}

    ⇒−2b2=−2bc2−a2x2[∵a2+b2=c2]\Rightarrow -2b^2 = -2b\sqrt{c^2 - a^2x^2}[\because a^2 + b^2 = c^2]

    ⇒b=c2−a2x2⇒a2x2=c2−b2=a2⇒x=±1\Rightarrow b = \sqrt{c^2 - a^2x^2} \Rightarrow a^2x^2 = c^2 - b^2 = a^2 \Rightarrow x = \pm 1

    Clearly, x=0,±1x = 0, \pm 1 satisfy the equation.

  116. Let f(x)=sin⁡[2cos⁡−1{cot⁡(2tan⁡−1x)}]f(x) = \sin[2\cos^{-1}\{\cot(2\tan^{-1}x)\}]

    =sin⁡[2cos⁡−1{cot⁡tan⁡−12x1−x2}]= \sin\left[2\cos^{-1}\left\{\cot \tan^{-1}\frac{2x}{1 - x^2}\right\}\right]

    =sin⁡[2cos⁡−1(cot⁡cot⁡−11−x22x)]= \sin\left[2\cos^{-1}\left(\cot\cot^{-1}\frac{1 - x^2}{2x}\right)\right]

    =sin⁡[2cos⁡−11−x22x]= \sin\left[2\cos^{-1}\frac{1 - x^2}{2x}\right]

    =sin⁡sin⁡−1[2.1−x22x1−(1−x22x)2]= \sin\sin^{-1}\left[2.\frac{1 - x^2}{2x}\sqrt{1- \left(\frac{1 - x^2}{2x}\right)^2}\right]

    =1−x2x1−(1−x22x)2= \frac{1 - x^2}{x}\sqrt{1 - \left(\frac{1 - x^2}{2x}\right)^2}

    When f(x)=0,f(x) = 0, we have (1−x2)1−(1−x22x)2=0(1 - x^2)\sqrt{1 - \left(\frac{1 - x^2}{2x}\right)^2} = 0

    ⇒(1−x2)6x2−1−x4=0\Rightarrow (1 - x^2)\sqrt{6x^2 - 1 - x^4} = 0

    Either 1−x2=01 - x^2 = 0 or 6x2−1−x4=0\sqrt{6x^2 - 1 - x^4} = 0

    ⇒x=±1\Rightarrow x = \pm1 or x4−6x2+1=0x^4 - 6x^2 + 1 = 0

    x2=6±36−42=3±22=(1±2)2x^2 = \frac{6\pm \sqrt{36 - 4}}{2} = 3 \pm 2\sqrt{2} = (1 \pm \sqrt{2})^2

    x=±(1±2)x = \pm(1 \pm \sqrt{2})

    x=±1,±(1±2)x = \pm 1, \pm(1 \pm \sqrt{2})

  117. 3sin⁡2θ5+4cos⁡2θ=3.2tan⁡θ1+tan⁡2θ5+41−tan⁡2θ1+tan⁡2θ\frac{3\sin2\theta}{5 + 4\cos2\theta} = \frac{3.\frac{2\tan\theta}{1 + \tan^2\theta}}{5 + 4\frac{1 - \tan^2\theta}{1 + \tan^2\theta}}

    =6tan⁡θ9+tan⁡2θ=2tan⁡ϕ1+tan⁡2ϕ= \frac{6\tan\theta}{9 + \tan^2\theta} = \frac{2\tan\phi}{1 + \tan^2\phi} where 13tan⁡θ=tan⁡ϕ\frac{1}{3}\tan\theta = \tan\phi

    =sin⁡2ϕ= \sin2\phi

    Given equation is θ=tan⁡−1(2tan⁡2θ)−12sin⁡−1(3sin⁡2θ5+4cos⁡2θ)\theta = \tan^{-1}(2\tan^2\theta) - \frac{1}{2}\sin^{-1}\left(\frac{3\sin2\theta}{5 + 4\cos2\theta}\right)

    ⇒tan⁡−1(2tan⁡2θ)−12sin⁡−1sin⁡2ϕ=θ\Rightarrow \tan^{-1}(2\tan^2\theta) - \frac{1}{2}\sin^{-1}\sin2\phi = \theta

    ⇒tan⁡−1(2tan⁡2θ)−tan⁡−113tan⁡θ=θ\Rightarrow \tan^{-1}(2\tan^2\theta) - \tan^{-1}\frac{1}{3}\tan\theta = \theta

    ⇒tan⁡−1[2tan⁡2θ−13tan⁡θ1+2tan⁡2θ.13tan⁡θ]=θ\Rightarrow \tan^{-1}\left[\frac{2\tan^2\theta - \frac{1}{3}\tan\theta}{1 + 2\tan^2\theta.\frac{1}{3}\tan\theta}\right] = \theta

    ⇒tan⁡θ(2tan⁡θ−13)1+23tan⁡3θ=tan⁡θ\Rightarrow \frac{\tan\theta\left(2\tan\theta - \frac{1}{3}\right)}{1 + \frac{2}{3}\tan^3\theta} = \tan\theta

    ⇒tan⁡θ[6tan⁡θ−13+2tan⁡2θ−1]=0\Rightarrow \tan\theta\left[\frac{6\tan\theta - 1}{3 + 2\tan^2\theta} - 1\right] = 0

    If tan⁡θ=0⇒θ=nπ\tan\theta = 0 \Rightarrow \theta = n\pi

    If 6tan⁡θ−13+2tan⁡3θ−1=0\frac{6\tan\theta - 1}{3 + 2\tan^3\theta} - 1 = 0

    ⇒tan⁡3θ−3tan⁡θ+2=0⇒(tan⁡θ−1)2(tan⁡θ+2)=0\Rightarrow \tan^3\theta - 3\tan\theta + 2 = 0 \Rightarrow (\tan\theta - 1)^2(\tan\theta + 2) = 0

    Either tan⁡θ=1⇒θ=nπ+π4\tan\theta = 1 \Rightarrow \theta = n\pi + \frac{\pi}{4}

    or tan⁡θ=−2⇒θ=nπ+tan⁡−1(−2)\tan\theta = -2 \Rightarrow \theta = n\pi + \tan^{-1}(-2)

  118. tan⁡−12x+tan⁡−13x=π4\tan^{-1}2x + \tan^{-1}3x = \frac{\pi}{4}

    ⇒tan⁡−13x=π4−tan⁡−12x\Rightarrow \tan^{-1}3x = \frac{\pi}{4} - \tan^{-1}2x

    ⇒3x=tan⁡(π4−tan⁡−12x)\Rightarrow 3x = \tan\left(\frac{\pi}{4} - \tan^{-1}2x\right)

    3x=1−tan⁡(tan⁡−12x)1+tan⁡(tan⁡−12x)=1−2x1+2x3x = \frac{1 - \tan(\tan^{-1}2x)}{1 + \tan(\tan^{-1}2x)} = \frac{1 - 2x}{1 + 2x}

    ⇒6x2+3x=1−2x⇒6x2+5x−1=0\Rightarrow 6x^2 + 3x = 1 - 2x \Rightarrow 6x^2 + 5x - 1 = 0

    ⇒(x+1)(6x−1)=0\Rightarrow (x + 1)(6x - 1) = 0

    ⇒x=−1,16\Rightarrow x = -1, \frac{1}{6}

    When x=−1x = -1 L.H.S. is negative angle but R.H.S. is positive so it is not a solution. However, for x=16x = \frac{1}{6} satisfies both sides are positive and balanced.

  119. Given, sin⁡−1(x1+x2)+cos⁡−1(x21+x22)=π2\sin^{-1}\left(\frac{x}{1 + \frac{x}{2}}\right) + \cos^{-1}\left(\frac{x^2}{1 + \frac{x^2}{2}}\right) = \frac{\pi}{2}

    ⇒sin⁡−1(2xx+2)=π2−cos⁡−1(2x22+x2)=sin⁡−1(2x22+x2)\Rightarrow \sin^{-1}\left(\frac{2x}{x + 2}\right) = \frac{\pi}{2} - \cos^{-1}\left(\frac{2x^2}{2 + x^2}\right) = \sin^{-1}\left(\frac{2x^2}{2 + x^2}\right)

    ⇒2x2+x=2x22+x2\Rightarrow \frac{2x}{2 + x} = \frac{2x^2}{2 + x^2}

    ⇒2x[12+x−x2+x2]=0\Rightarrow 2x\left[\frac{1}{2 + x} - \frac{x}{2 + x^2}\right] = 0

    ⇒x=0\Rightarrow x = 0 or 2+x2=2x+x2⇒x=12 + x^2 = 2x + x^2 \Rightarrow x = 1

    But 0<∣x∣<2⇒x=10 < |x| < \sqrt{2} \Rightarrow x = 1

  120. Given, tan⁡−1x(x+1)+sin⁡−1x2+x+1=π2\tan^{-1}\sqrt{x(x + 1)} + \sin^{-1}\sqrt{x^2 + x + 1} = \frac{\pi}{2}

    ⇒tan⁡−1x(x+1)+tan⁡−1x2+x+1x(x+1)=π2\Rightarrow \tan^{-1}\sqrt{x(x + 1)} + \tan^{-1}\frac{\sqrt{x^2 + x + 1}}{\sqrt{x(x + 1)}} = \frac{\pi}{2}

    tan⁡−1[x(x+1)+x2+x+1x(x+1)1−x(x+1).x2+x+1x(x+1)]=π2\tan^{-1}\left[\frac{\sqrt{x(x + 1)} + \frac{\sqrt{x^2 + x + 1}}{\sqrt{x(x + 1)}}}{1 - \sqrt{x(x + 1)}.\frac{\sqrt{x^2 + x + 1}}{\sqrt{x(x + 1)}}}\right] = \frac{\pi}{2}

    ⇒1−x2+x+1=0\Rightarrow 1 - \sqrt{x^2 + x + 1} = 0

    ⇒x=0,−1\Rightarrow x = 0, -1

    Clearly both these values of xx satisfy the equation.

  121. This problem is similar to 115 so x=−1,0,1x = -1, 0, 1.

  122. We have to solve sin⁡−1(1−x)−2sin⁡−1x=π2\sin^{-1}(1 - x) - 2\sin^{-1}x = \frac{\pi}{2}

    Let x=sin⁡yx = \sin y

    ⇒sin⁡−1(1−sin⁡y)−2y=π2\Rightarrow \sin^{-1}(1 - \sin y) - 2y = \frac{\pi}{2}

    ⇒sin⁡−1(1−sin⁡y)=π2+2y\Rightarrow \sin^{-1}(1 - \sin y) =\frac{\pi}{2} + 2y

    ⇒1−sin⁡y=sin⁡(π2+2y)=cos⁡2y=1−2sin⁡2y\Rightarrow 1 - \sin y = \sin\left(\frac{\pi}{2} + 2y\right) = \cos2y = 1 - 2\sin^2y

    ⇒2sin⁡2y−sin⁡y=0\Rightarrow 2\sin^2y - \sin y = 0

    ⇒2x2−x=0\Rightarrow 2x^2 - x = 0

    x=0,12x = 0, \frac{1}{2}

    But x=12x = \frac{1}{2} does not satisfy the equation but x=0x = 0 does so it is the required solution.

  123. Given equation is tan⁡−1x+tan⁡−1y=tan⁡−1k\tan^{-1}x + \tan^{-1}y = \tan^{-1}k

    ⇒tan⁡−1x+y1−xy=tan⁡−1k\Rightarrow \tan^{-1}\frac{x + y}{1 - xy} = \tan^{-1}k

    ⇒x+y1−xy=k\Rightarrow \frac{x + y}{1 - xy} = k

    For k>0,1−xy>0⇒xy<1k > 0, 1 - xy > 0 \Rightarrow xy < 1 which implies both xx and yy cannot be positive integers.

  124. We have to solve tan⁡−1x+1x−1+tan⁡−1x−1x=tan⁡−1(−7)\tan^{-1}\frac{x + 1}{x - 1} + \tan^{-1}\frac{x - 1}{x} = \tan^{-1}(-7)

    ⇒x+1x−1+x−1x1−x+1x−1.x−1x=−7\Rightarrow \frac{\frac{x + 1}{x - 1} + \frac{x - 1}{x}}{1 - \frac{x + 1}{x - 1}.\frac{x - 1}{x}} = -7

    ⇒x2+x+x2−2x+1x2−x−x2+1=−7\Rightarrow \frac{x^2 + x + x^2 - 2x + 1}{x^2 - x - x^2 + 1} = -7

    ⇒2x2−x+1=7x−7\Rightarrow 2x^2 - x + 1 = 7x - 7

    ⇒2x2−8x+8=0⇒x2−4x+4=0⇒x=2\Rightarrow 2x^2 - 8x + 8 = 0 \Rightarrow x^2 - 4x + 4 = 0 \Rightarrow x = 2

  125. We have to solve tan⁡−11a−1=tan⁡−11x+tan⁡−11a2−x+1\tan^{-1}\frac{1}{a - 1} = \tan^{-1}\frac{1}{x} + \tan^{-1}\frac{1}{a^2 - x + 1}

    ⇒1a−1=a2+1a2x−x2+x−1\Rightarrow \frac{1}{a - 1} = \frac{a^2 + 1}{a^2x - x^2 + x - 1}

    ⇒(a−x)(a2−a−x+1)=0\Rightarrow (a - x)(a^2 - a - x + 1) = 0

    x=a,a2−a+1x = a, a^2 - a + 1

  126. We have to solve cos⁡−1x2−1x2+1+tan⁡−12xx2−1=2π3\cos^{-1}\frac{x^2 - 1}{x^2 + 1} + \tan^{-1}\frac{2x}{x^2 - 1} = \frac{2\pi}{3}

    Case I: When x>1x > 1 given equation becomes π−2tan⁡−1x+π−2tan⁡−1x=2π3\pi - 2\tan^{-1}x + \pi - 2\tan^{-1}x= \frac{2\pi}{3}

    ⇒tan⁡−1x=π3⇒x=3\Rightarrow \tan^{-1}x = \frac{\pi}{3} \Rightarrow x = \sqrt{3}

    Case II: When x<1x < 1 given equation becomes π−2tan⁡−1x−2tan⁡−1x=2π3\pi - 2\tan^{-1}x - 2\tan^{-1}x = \frac{2\pi}{3}

    ⇒x=tan⁡π12=3−13+1=2−3\Rightarrow x = \tan\frac{\pi}{12} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3}

  127. Given θ=tan⁡−1x32k−x\theta = \tan^{-1}\frac{x\sqrt{3}}{2k - x} and ϕ=tan⁡−12x−kk3\phi = \tan^{-1}\frac{2x - k}{k\sqrt{3}}.

    θ−ϕ=tan⁡−1x32k−x−2x−kk31+xk.2x−k2k−x\theta - \phi = \tan^{-1}\frac{\frac{x\sqrt{3}}{2k - x} - \frac{2x - k}{k\sqrt{3}}}{1 + \frac{x}{k}.\frac{2x - k}{2k - x}}

    =tan⁡−113.2k2+2x2−2kx2k2+2x2−2kx= \tan^{-1}\frac{1}{\sqrt{3}}.\frac{2k^2 + 2x^2 - 2kx}{2k^2 + 2x^2 - 2kx}

    If 2x2+2k2−2kx≠02x^2 + 2k^2 - 2kx\neq 0 then θ−ϕ=tan⁡−113=π6\theta - \phi = \tan^{-1}\frac{1}{\sqrt{3}} = \frac{\pi}{6}

  128. Given tan⁡−1x+cos⁡−1y1+y2=sin⁡−1310\tan^{-1}x + \cos^{-1}\frac{y}{\sqrt{1 + y^2}} = \sin^{-1}\frac{3}{\sqrt{10}}

    ⇒tan⁡−11y=tan⁡−13−tan⁡−1x\Rightarrow \tan^{-1}\frac{1}{y} = \tan^{-1}3 - \tan^{-1}x

    ⇒1y=3−x1+3x\Rightarrow \frac{1}{y} = \frac{3 - x}{1 + 3x}

    y=1+3x3−xy = \frac{1 + 3x}{3 - x}

    Clearly, for x≥3x \geq 3 there can be no solution as yy becomes infinity and negative for those values.

    When x=1,y=2x = 1, y = 2 and when x=2,y=7x = 2, y = 7.

  129. Given equation is an identity except for range. For sin⁡−1s1−x2\sin^{-1}s\sqrt{1 - x^2} range is [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

    For 2cos⁡−1x2\cos^{-1}x range is [0,2π][0, 2\pi] so common range is [0,π2]\left[0, \frac{\pi}{2}\right]

    ⇒0≤2cos⁡−1x≤π2\Rightarrow 0 \leq 2\cos^{-1}x \leq \frac{\pi}{2}

    ⇒12≤x≤1\Rightarrow \frac{1}{\sqrt{2}}\leq x \leq 1, since cos⁡\cos is a decreasing function, so inequality is reversed.

  130. We have to solve sin⁡−1x1+x2−sin⁡−111+x2=sin⁡−11+x1+x2\sin^{-1}\frac{x}{\sqrt{1 + x^2}} - \sin^{-1}\frac{1}{\sqrt{1 + x^2}} = \sin^{-1}\frac{1 + x}{1 + x^2}

    ⇒sin⁡−1x2−11+x2=sin⁡−11+x1+x2\Rightarrow \sin^{-1}\frac{x^2 - 1}{1 + x^2} = \sin^{-1}\frac{1 + x}{1 + x^2}

    ⇒x2−x−2=0⇒x=−1,2\Rightarrow x^2 - x - 2 = 0 \Rightarrow x = -1, 2

    However, the equation is not satisfied for x=−1x = -1 hence x=2x = 2 is the required solution.

  131. Given equation is y=2tan⁡−1[a−ba+btan⁡x2]−cos⁡−1[b+acos⁡xa+bcos⁡x]y = 2\tan^{-1}\left[\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}\right] - \cos^{-1}\left[\frac{b + a\cos x}{a + b\cos x}\right]

    R.H.S. =tan⁡−12a−ba+btan⁡x21−a−ba+btan⁡2x2−tan⁡−1a2−b2sin⁡xb+acos⁡x= \tan^{-1}\frac{2\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}}{1 - \frac{a - b}{a + b}\tan^2\frac{x}{2}} - \tan^{-1}\frac{\sqrt{a^2 - b^2}\sin x}{b + a\cos x}

    =tan⁡−12a2−b2tan⁡x2(a+b)−(a−b)tan⁡2x2−tan⁡−12a2−b2tan⁡x2(a+b)−(a−b)tan⁡2x2= \tan^{-1}\frac{2\sqrt{a^2 - b^2}\tan\frac{x}{2}}{(a + b) - (a - b)\tan^2\frac{x}{2}} - \tan^{-1}\frac{2\sqrt{a^2 - b^2}\tan\frac{x}{2}}{(a + b) - (a - b)\tan^2\frac{x}{2}}

    =0= 0 which is a constant provided a≥b>0a \geq b > 0

  132. tan−12i2+i2+i4=tan−1(i2+i+1)−tan⁡−1(i2−i+1)tan^{-1}\frac{2i}{2 + i^2 + i^4} = \\tan^{-1}(i^2 + i + 1) - \tan^{-1}(i^2 - i + 1)

    Subtituting i=1,i = 1, R.H.S. = tan^{-1}3 - tan^{-1}1

    Subtituting i=2,i = 2, R.H.S. = tan^{-1}7 - tan^{-1}3

    Subtituting i=3,i = 3, R.H.S. = tan^{-1}13 - tan^{-1}7

    …\ldots

    Subtituting i=n−1,i = n - 1, R.H.S. = tan^{-1}(n^2 - n + 1) - tan^{-1}(n^2 - 3n + 3)

    Subtituting i=n,i = n, R.H.S. = tan^{-1}(n^2 + n + 1) - tan^{-1}(n^2 - n + 1)

    Adding, we get ∑i=1ntan⁡−12i2+i2+i4=tan⁡−1(n2+n+1)−tan⁡−11\sum_{i=1}^{n}\tan^{-1}\frac{2i}{2 + i^2 + i^4} = \tan^{-1}(n^2 + n + 1) - \tan^{-1}1

    =tan⁡−1n2+nn2+n+2= \tan^{-1}\frac{n^2 + n}{n^2 + n + 2}

  133. tn=cot⁡−1(n2+34)=cot⁡−1(4n2+34)t_n = \cot^{-1}\left(n^2 + \frac{3}{4}\right) = \cot^{-1}\left(\frac{4n^2 + 3}{4}\right)

    =tan⁡−111+n4−14=(n+12)−(n−12)1+(n+12)(n−12)= \tan^{-1}\frac{1}{1 + n^4 - \frac{1}{4}} = \frac{\left(n + \frac{1}{2}\right) - \left(n - \frac{1}{2}\right)}{1 + \left(n + \frac{1}{2}\right)\left(n - \frac{1}{2}\right)}

    =tan⁡−1(n+12)−tan⁡−1(n−12)= \tan^{-1}\left(n + \frac{1}{2}\right) - \tan^{-1}\left(n - \frac{1}{2}\right)

    Putting n=1,t1=tan⁡−132−tan⁡−112n = 1, t_1 = \tan^{-1}\frac{3}{2} - \tan^{-1}\frac{1}{2}

    Putting n=2,t2=tan⁡−152−tan⁡−132n = 2, t_2 = \tan^{-1}\frac{5}{2} - \tan^{-1}\frac{3}{2}

    …\ldots

    Putting n=∞,t∞=tan⁡−1(∞+12)−tan⁡−1(∞−12)n = \infty, t_\infty = \tan^{-1}\left(\infty + \frac{1}{2}\right) - \tan^{-1}\left(\infty - \frac{1}{2}\right)

    Adding, we get S=tan⁡−1(∞+12)−tan⁡−112S = \tan^{-1}\left(\infty + \frac{1}{2}\right) - \tan^{-1}\frac{1}{2}

    =π2−tan⁡−112=cot⁡−112=tan⁡−12= \frac{\pi}{2} - \tan^{-1}\frac{1}{2} = \cot^{-1}\frac{1}{2} = \tan^{-1}2

  134. We know that tan⁡−1x+cot⁡−1x=π2\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}

    So given equation can be written as (tan⁡−1x+cot⁡−1x)−2tan⁡−1x(π2−tan⁡−1x)=5π28(\tan^{-1}x + \cot^{-1}x) - 2\tan^{-1}x\left(\frac{\pi}{2} - \tan^{-1}x\right) = \frac{5\pi^2}{8}

    ⇒2(tan⁡−1x)−πtan⁡−1x−3π28=0\Rightarrow 2(\tan^{-1}x) - \pi\tan^{-1}x - \frac{3\pi^2}{8} = 0

    ⇒tan⁡−1x=−π4,3π4⇒x=−1\Rightarrow \tan^{-1}x = -\frac{\pi}{4}, \frac{3\pi}{4} \Rightarrow x = -1

  135. (sin⁡−1x+cos⁡−1x)3=(sin⁡−1x+cos⁡−1x)[(sin⁡−1x+cos⁡−1x)2−3sin⁡−1xcos⁡−1x](\sin^{-1}x + \cos^{-1}x)^3 = (\sin^{-1}x + \cos^{-1}x)[(\sin^{-1}x + \cos^{-1}x)^2 - 3\sin^{-1}x\cos^{-1}x]

    =π2[π24−3sin⁡−1x(π2−sin⁡−1x)]= \frac{\pi}{2}\left[\frac{\pi^2}{4} - 3\sin^{-1}x\left(\frac{\pi}{2} - \sin^{-1}x\right)\right]

    =3π2[(sin⁡−1x−π4)2+π248]= \frac{3\pi}{2}\left[\left(\sin^{-1}x - \frac{\pi}{4}\right)^2 + \frac{\pi^2}{48}\right]

    Least value of sin⁡−1−π4=0\sin^{-1} - \frac{\pi}{4} = 0 and greatest value is 3π4\frac{3\pi}{4}

    Hence greatest and least values of the required expression are 7π38\frac{7\pi^3}{8} and π232\frac{\pi^2}{32}.

  136. Let cos⁡−1x=a∈[0,π]\cos^{-1}x = a \in [0, \pi] and sin⁡−1y=b∈[−π2,π2]\sin^{-1}y = b \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

    We have a+b2=pπ24a + b^2 = \frac{p\pi^2}{4} and ab2=π416ab^2 = \frac{\pi^4}{16}

    Since b2∈[0,π2/4]b^2\in [0, \pi^2/4], we get a+b2∈[0,π+π2/4]a + b^2 \in [0, \pi + \pi^2/4]

    ⇒0≤pπ24≤π+π24\Rightarrow 0 \leq \frac{p\pi^2}{4}\leq \pi + \frac{\pi^2}{4}

    ⇒0≤p≤4π+1\Rightarrow 0\leq p\leq \frac{4}{\pi} + 1

    So p=0,1,2p = 0, 1, 2

    ⇒a(pπ24−a)=π216\Rightarrow a\left(\frac{p\pi^2}{4} - a\right) = \frac{\pi^2}{16}

    Since a∈R⇒D≥0a\in R \Rightarrow D\geq 0

    p2≥4⇒p=2p^2 \geq 4\Rightarrow p = 2

  137. Given tan⁡−1x,tan⁡−1y,tan⁡−1z\tan^{-1}x, \tan^{-1}y, \tan^{-1}z are in A.P.

    ⇒2tan⁡−1y=tan⁡−1x+tan⁡−1z\Rightarrow 2\tan^{-1}y = \tan^{-1}x + \tan^{-1}z

    ⇒2y1−y2=x+z1−xz\Rightarrow \frac{2y}{1 - y^2} = \frac{x + z}{1 - xz}

    If x,y,zx, y, z are in A.P. 2y=x+z2y = x + z

    ⇒1−(x+z)24=1−xz\Rightarrow 1 - \frac{(x + z)^2}{4} = 1 - xz

    ⇒(x−z)2=0⇒x=y=z\Rightarrow (x - z)^2 = 0 \Rightarrow x = y = z

  138. tn=tan⁡−1x1+n(n+1)x2=tan⁡−1(n+1)x−tan⁡−1nxt_n = \tan^{-1}\frac{x}{1 + n(n + 1)x^2} = \tan^{-1}(n + 1)x - \tan^{-1}nx

    t1=tan⁡−12x−tan⁡−1xt_1 = \tan^{-1}2x - \tan^{-1}x

    t2=tan⁡−13x−tan⁡−12xt_2 = \tan^{-1}3x - \tan^{-1}2x

    …\ldots

    tn=tan⁡−1(n+1)x−tan⁡−1nxt_n = \tan^{-1}(n + 1)x - \tan^{-1}nx

    Adding, we get S=tan⁡−1(n+1)x−tan⁡−1x=tan⁡−1nx1+(n+1)x2=S = \tan^{-1}(n + 1)x - \tan^{-1}x = \tan^{-1}\frac{nx}{1 + (n + 1)x^2} = R.H.S.