9. Compound Angles Solutions#

  1. Given, sin⁡α=35\sin\alpha = \frac{3}{5} and cos⁡β=941\cos\beta = \frac{9}{41}

    Therefore, cos⁡α=45\cos\alpha = \frac{4}{5} and sin⁡β=4041\sin\beta = \frac{40}{41}

    sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta

    =35941−454041= \frac{3}{5}\frac{9}{41} - \frac{4}{5}\frac{40}{41}

    =27−160205=−133205= \frac{27 - 160}{205} = -\frac{133}{205}

    cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta

    =45941−354041= \frac{4}{5}\frac{9}{41} - \frac{3}{5}\frac{40}{41}

    =36−120205=−84205= \frac{36 - 120}{205} = -\frac{84}{205}

  2. Given, sin⁡α=4553\sin\alpha = \frac{45}{53} and sin⁡β=3365\sin\beta = \frac{33}{65}

    Thus, cos⁡α=2853\cos\alpha = \frac{28}{53} and cos⁡β=5665\cos\beta = \frac{56}{65}

    sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta

    =45535665−28533365= \frac{45}{53}\frac{56}{65} - \frac{28}{53}\frac{33}{65}

    =2520−9243445=15963445= \frac{2520 - 924}{3445} = \frac{1596}{3445}

    sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta

    =45535665+28533365= \frac{45}{53}\frac{56}{65} + \frac{28}{53}\frac{33}{65}

    =2520+9243445=34443445= \frac{2520 + 924}{3445} = \frac{3444}{3445}

  3. Given, sin⁡α=1517\sin\alpha = \frac{15}{17} and cos⁡β=1213\cos\beta = \frac{12}{13}

    cos⁡α=817\cos\alpha = \frac{8}{17} and sin⁡β=513\sin\beta = \frac{5}{13}

    tan⁡α=158\tan\alpha = \frac{15}{8} and tan⁡β=512\tan\beta = \frac{5}{12}

    sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta

    =15171213+817513= \frac{15}{17}\frac{12}{13} + \frac{8}{17}\frac{5}{13}

    =220221= \frac{220}{221}

    cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta

    =8171213+1517513= \frac{8}{17}\frac{12}{13} + \frac{15}{17}\frac{5}{13}

    =171221= \frac{171}{221}

    tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}

    =158+5121−158512= \frac{\frac{15}{8} + \frac{5}{12}}{1 - \frac{15}{8}\frac{5}{12}}

    =22021= \frac{220}{21}

  4. L.H.S. =cos⁡(45∘−A)cos⁡(45∘−B)−sin⁡(45∘−A)sin⁡(45∘−B)= \cos(45^{\circ} - A)\cos(45^{\circ} - B) - \sin(45^{\circ} - A)\sin(45^{\circ} - B)

    =[(cos⁡45∘cos⁡A+sin⁡45∘sin⁡A)(cos⁡45∘cos⁡B+sin⁡45∘sin⁡B)−(sin⁡45∘cos⁡A−cos⁡45∘sin⁡A)(sin⁡45∘cos⁡B−cos⁡45∘sin⁡B)]= [(\cos 45^\circ\cos A + \sin45^\circ\sin A)(\cos 45^\circ\cos B + \sin45^\circ\sin B) - (\sin45^\circ\cos A - \cos45^\circ\sin A)(\sin45^\circ\cos B - \cos45^\circ\sin B)]

    Substituting valus for sin⁡45∘\sin45^\circ and cos⁡45∘\cos45^\circ

    =[(cos⁡A2+sin⁡A2)(cos⁡B2+sin⁡B2)]−[(cos⁡A2−sin⁡A2)(cos⁡B2−sin⁡B2)]=\left[\left(\frac{\cos A}{\sqrt{2}} + \frac{\sin A}{\sqrt{2}}\right)\left(\frac{\cos B}{\sqrt{2}} + \frac{\sin B}{\sqrt{2}}\right)\right] - \left[\left(\frac{\cos A}{\sqrt{2}} - \frac{\sin A}{\sqrt{2}}\right)\left(\frac{\cos B}{\sqrt{2}} - \frac{\sin B}{\sqrt{2}}\right)\right]

    =[cos⁡Acos⁡B2+cos⁡Asin⁡B2+sin⁡Acos⁡B2+sin⁡Asin⁡B2]−[cos⁡Acos⁡B2−cos⁡Asin⁡B2−sin⁡Acos⁡B2+sin⁡Asin⁡B2]= \left[\frac{\cos A\cos B}{2} + \frac{\cos A\sin B}{2} + \frac{\sin A\cos B}{2} + \frac{\sin A\sin B}{2}\right] - \left[\frac{\cos A\cos B}{2} - \frac{\cos A\sin B}{2} - \frac{\sin A\cos B}{2} + \frac{\sin A\sin B}{2}\right]

    =sin⁡Acos⁡B+cos⁡Asin⁡B=sin⁡(A+B)= \sin A\cos B + \cos A\sin B = \sin(A + B)

  5. L.H.S. =sin⁡(45∘+A)cos⁡(45∘−B)+cos⁡(45∘+A)sin⁡(45∘−B)= \sin(45^{\circ} + A)\cos(45^\circ - B) + \cos(45^{\circ} + A)\sin(45^\circ - B)

    =[(sin⁡45∘cos⁡A+cos⁡45∘sin⁡A)(cos⁡45∘cos⁡B+sin⁡45∘sin⁡B)+(cos⁡45∘cos⁡A−sin⁡45∘sin⁡A)(sin⁡45∘cos⁡B−cos⁡45∘)sin⁡B]= [(\sin45^\circ\cos A + \cos45^\circ\sin A)(\cos45^\circ\cos B + \sin45^\circ\sin B) + (\cos45^\circ\cos A - \sin45^\circ\sin A)(\sin45^\circ\cos B - \cos45^\circ)\sin B]

    =[(cos⁡A2+sin⁡A2)(cos⁡B2+sin⁡B2)+(cos⁡A2−sin⁡A2)(cos⁡B2−sin⁡B2)]= \left[\left(\frac{\cos A}{\sqrt{2}} + \frac{\sin A}{\sqrt{2}}\right)\left(\frac{\cos B}{\sqrt{2}} + \frac{\sin B}{\sqrt{2}}\right) + \left(\frac{\cos A}{\sqrt{2}} - \frac{\sin A}{\sqrt{2}}\right)\left(\frac{\cos B}{\sqrt{2}} - \frac{\sin B}{\sqrt{2}}\right)\right]

    =[cos⁡Acos⁡B2+cos⁡Asin⁡B2+sin⁡Acos⁡B2+sin⁡Asin⁡B2+cos⁡Acos⁡B2−cos⁡Asin⁡B2−sin⁡Acos⁡B2+sin⁡Asin⁡B2]= \left[\frac{\cos A\cos B}{2} + \frac{\cos A\sin B}{2} + \frac{\sin A\cos B}{2} + \frac{\sin A\sin B}{2} + \frac{\cos A\cos B}{2} - \frac{\cos A\sin B}{2} - \frac{\sin A\cos B}{2} + \frac{\sin A\sin B}{2}\right]

    =cos⁡Acos⁡B+sin⁡Asin⁡B=cos⁡(A−B)= \cos A\cos B + \sin A\sin B = \cos(A - B)

  6. L.H.S. =sin⁡(A−B)cos⁡Acos⁡B+sin⁡(B−C)cos⁡Bcos⁡C+sin⁡(C−A)cos⁡Ccos⁡A= \frac{\sin(A - B)}{\cos A\cos B} + \frac{\sin(B - C)}{\cos B\cos C} + \frac{\sin(C - A)}{\cos C\cos A}

    =sin⁡Acos⁡B−cos⁡Asin⁡Bcos⁡Acos⁡B+sin⁡Bcos⁡C−cos⁡Bsin⁡Ccos⁡Bcos⁡C+sin⁡Ccos⁡A−cos⁡Csin⁡Acos⁡Ccos⁡A= \frac{\sin A\cos B - \cos A\sin B}{\cos A\cos B} + \frac{\sin B\cos C - \cos B\sin C}{\cos B\cos C} + \frac{\sin C\cos A - \cos C\sin A}{\cos C\cos A}

    =tan⁡A−tan⁡B+tan⁡B−tan⁡C+tan⁡C−tan⁡A=0== \tan A - \tan B + \tan B - \tan C + \tan C - \tan A = 0 = R.H.S.

  7. L.H.S. =sin⁡105∘+cos⁡105∘= \sin 105^\circ + \cos 105^\circ

    =sin⁡(60∘+45∘)+cos⁡(60∘+45∘)= \sin(60^\circ + 45^\circ) + \cos(60^\circ + 45^\circ)

    =sin⁡60∘cos⁡45∘+cos⁡60∘sin⁡45∘+cos⁡60∘cos⁡45∘−sin⁡60∘sin⁡45∘=\sin60^\circ\cos45^\circ + \cos60^\circ\sin45^\circ + \cos60^\circ\cos45^\circ - \sin60^\circ\sin45^\circ

    =cos⁡45∘(sin⁡60∘+cos⁡60∘+cos⁡60∘−sin⁡60∘)[∵sin⁡45∘=cos⁡45∘]=\cos45^\circ(\sin60^\circ + \cos60^\circ + \cos60^\circ - \sin 60^\circ)[\because \sin45^\circ = \cos45^\circ]

    =cos⁡45∘== \cos45^\circ = R.H.S.

  8. Given, sin⁡75∘−sin⁡15∘=cos⁡105∘+cos⁡15∘\sin 75^\circ - \sin 15^\circ = \cos 105^\circ + \cos 15^\circ

    ⇒sin⁡75∘−sin⁡15∘=cos⁡(90∘+15∘)+sin⁡(90∘−15∘)\Rightarrow \sin75^\circ -\sin15^\circ = \cos(90^\circ + 15^\circ) + \sin(90^\circ - 15^\circ)

    ⇒sin⁡75∘−sin⁡15∘=cos⁡90∘cos⁡15∘−sin⁡90∘sin⁡15∘+sin⁡(90∘−15∘)\Rightarrow \sin75^\circ -\sin15^\circ = \cos90^\circ\cos15^\circ - \sin90^\circ\sin15^\circ + \sin(90^\circ - 15^\circ)

    ⇒sin⁡75∘−sin⁡15∘=−sin⁡15∘+sin⁡75∘[∵cos⁡90∘=0 & sin⁡90∘=1]\Rightarrow \sin75^\circ -\sin15^\circ = -\sin15^\circ + \sin75^\circ [\because \cos90^\circ = 0~\&~\sin90^\circ = 1]

    Thus, we have proven the equality.

  9. L.H.S. =cos⁡αcos⁡(γ−α)−sin⁡αsin⁡(γ−α)= \cos\alpha\cos(\gamma - \alpha) - \sin\alpha\sin(\gamma - \alpha)

    =cos⁡α(cos⁡γcos⁡α+sin⁡γsin⁡α)−sin⁡α(sin⁡γcos⁡α−sin⁡αcos⁡γ)= \cos\alpha(\cos\gamma\cos\alpha + \sin\gamma\sin\alpha) - \sin\alpha(\sin\gamma\cos\alpha - \sin\alpha\cos\gamma)

    =cos⁡2αcos⁡γ+sin⁡γsin⁡αcos⁡α−sin⁡αsin⁡γcos⁡α−sin⁡2αcos⁡γ= \cos^2\alpha\cos\gamma + \sin\gamma\sin\alpha\cos\alpha - \sin\alpha\sin\gamma\cos\alpha - \sin^2\alpha\cos\gamma

    =cos⁡γ(sin⁡2α+cos⁡2α)=cos⁡γ== \cos\gamma(\sin^2\alpha + \cos^2\alpha) = \cos\gamma = R.H.S.

  10. L.H.S. =cos⁡(α+β)cos⁡γ−cos⁡(β+γ)cos⁡α= \cos(\alpha + \beta)\cos\gamma - \cos(\beta + \gamma)\cos\alpha

    =cos⁡αcos⁡βcos⁡γ−sin⁡αsin⁡βcos⁡γ−cos⁡αcos⁡βcos⁡γ+cos⁡αsin⁡βsin⁡γ= \cos\alpha\cos\beta\cos\gamma - \sin\alpha\sin\beta\cos\gamma - \cos\alpha\cos\beta\cos\gamma + \cos\alpha\sin\beta\sin\gamma

    =sin⁡β(cos⁡αsin⁡γ−sin⁡αcos⁡γ)= \sin\beta(\cos\alpha\sin\gamma - \sin\alpha\cos\gamma)

    =sin⁡βsin⁡(γ−α)== \sin\beta\sin(\gamma - \alpha) = R.H.S.

  11. L.H.S. =sin⁡(n+1)Asin⁡(n−1)A+cos⁡(n+1)Acos⁡(n−1)A= \sin(n + 1)A\sin(n - 1)A + \cos(n + 1)A\cos(n - 1)A

    =cos⁡(n+1−(n−1))A=cos⁡2A== \cos(n + 1 - (n - 1))A = \cos 2A = R.H.S.

  12. L.H.S. =sin⁡(n+1)Asin⁡(n+2)A+cos⁡(n+1)Acos⁡(n+2)A= \sin(n + 1)A\sin(n + 2)A + \cos(n + 1)A\cos(n + 2)A

    =cos⁡(n+1−(n+1))=cos⁡A== \cos(n + 1 - (n + 1)) = \cos A = R.H.S.

  13. cos⁡15∘=cos⁡(45∘−cos⁡30∘)\cos 15^\circ = \cos(45^\circ - \cos 30^\circ)

    =cos⁡45∘cos⁡30∘+sin⁡45∘sin⁡30∘= \cos45^\circ\cos30^\circ + \sin45^\circ\sin30^\circ

    =1232+1212= \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\frac{1}{2}

    =3+122= \frac{\sqrt{3} + 1}{2\sqrt{2}}

    sin⁡105∘=sin⁡(60∘+45∘)\sin 105^\circ = \sin(60^\circ + 45^\circ)

    =sin⁡60∘cos⁡45∘+cos⁡60∘sin⁡45∘= \sin60^\circ\cos^45\circ + \cos60^\circ\sin45^\circ

    =3212+1212= \frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}} + \frac{1}{2}\frac{1}{\sqrt{2}}

    =3+122== \frac{\sqrt{3} + 1}{2\sqrt{2}} = R.H.S.

  14. tan⁡105∘=tan⁡(60∘+45∘)\tan 105^\circ = \tan(60^\circ + 45^\circ)

    =tan⁡60∘+tan⁡45∘1−tan⁡60∘tan⁡45∘= \frac{\tan60^\circ + \tan 45^\circ}{1 - \tan60^\circ\tan 45^\circ}

    =3+11−3= \frac{\sqrt{3} + 1}{1 - \sqrt{3}}

  15. tan⁡495∘cot⁡855∘=tan⁡(360∘+135∘)cot⁡(720∘+135∘)\frac{\tan 495^\circ}{\cot 855^\circ} = \frac{\tan(360^\circ + 135^\circ)}{\cot(720^\circ + 135^\circ)}

    =tan⁡135∘cot⁡135∘=tan⁡2135∘=(−1)2=1= \frac{\tan 135^\circ}{\cot 135^\circ} = \tan^2135^\circ = (-1)^2 = 1

  16. sin⁡(π+θ)=−sin⁡θ∴sin⁡(nπ+θ)=(1)nsin⁡θ\sin(\pi + \theta) = -\sin\theta \therefore \sin(n\pi + \theta) = (1)^n\sin\theta

    sin⁡(nπ+(−1)nπ4)=(−1)nsin⁡((−1)nπ4)\sin\left(n\pi + (-1)^n \frac{\pi}{4}\right) = (-1)^n\sin\left((-1)^n\frac{\pi}{4}\right)

    =(−1)n(−1)nsin⁡π4 [∵sin⁡(−θ)=−sin⁡θ]= (-1)^n(-1)^n\sin \frac{\pi}{4}~[\because \sin(-\theta) = -\sin\theta]

    =(−1)2nsin⁡π4=12= (-1)^{2n}\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}

  17. L.H.S. =sin⁡15∘=sin⁡(60∘−45∘)= \sin 15^\circ = \sin(60^\circ - 45^\circ)

    =sin⁡60∘cos⁡45∘−cos⁡60∘sin⁡45∘= \sin60^\circ\cos45^\circ - \cos60^\circ\sin45^\circ

    =3212−1212= \frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}} - \frac{1}{2}\frac{1}{\sqrt{2}}

    =3−122== \frac{\sqrt{3} - 1}{2\sqrt{2}} = R.H.S.

  18. L.H.S. =cos⁡75∘=cos⁡(45∘+30∘)= \cos 75^\circ = \cos(45^\circ + 30^\circ)

    =cos⁡45∘cos⁡30∘−sin⁡45∘sin⁡30∘= \cos45^\circ\cos30^\circ - \sin45^\circ\sin30^\circ

    =1232−1212= \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\frac{1}{2}

    =3−122== \frac{\sqrt{3} - 1}{2\sqrt{2}} = R.H.S.

  19. L.H.S. =tan⁡75∘=tan⁡(45∘+30∘)= \tan 75^\circ = \tan(45^\circ + 30^\circ)

    =tan⁡45∘+tan⁡30∘1−tan⁡45∘tan⁡30∘= \frac{\tan45^\circ + \tan30^\circ}{1 - \tan45^\circ\tan30^\circ}

    =1+131−1.13= \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1.\frac{1}{\sqrt{3}}}

    =3+133−13= \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}}

    =3+13−1=(3+1)23−1=2+3== \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = 2 + \sqrt{3} = R.H.S.

  20. L.H.S. =tan⁡15∘=tan⁡(45∘−30∘)= \tan 15^\circ = \tan(45^\circ - 30^\circ)

    =tan⁡45∘−tan⁡30∘1+tan⁡45∘tan⁡30∘= \frac{\tan45^\circ - \tan30^\circ}{1 + \tan45^\circ\tan30^\circ}

    =1−131+1.13= \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1.\frac{1}{\sqrt{3}}}

    =3−133+13= \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}}

    =3−13+1=(3−1)23−1=2−3== \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = 2 - \sqrt{3} = R.H.S.

  21. cos⁡1395∘=cos⁡(3∗360∘+315∘)=cos⁡315∘=cos⁡(270∘+45∘)\cos 1395^\circ = \cos(3*360^\circ + 315^\circ) = \cos315^\circ = \cos(270^\circ + 45^\circ)

    =cos⁡45∘=12= \cos45^\circ = \frac{1}{\sqrt{2}}

  22. tan⁡(−330∘)=−tan⁡(330∘)=−tan⁡(270∘+60∘)==cot⁡60∘=13\tan(-330^\circ) = -\tan(330^\circ) = -\tan(270^\circ + 60^\circ) = =\cot 60^\circ = \frac{1}{\sqrt{3}}

  23. Given, sin⁡300∘cosec⁡1050∘−tan⁡(−120∘)\sin 300^\circ \cosec 1050^\circ - \tan(-120^\circ)

    =sin⁡(270∘+30∘)cosec⁡(720∘+270∘+60∘)+tan⁡(90∘+30∘)= \sin (270^\circ + 30^\circ)\cosec(720^\circ + 270^\circ + 60^\circ) + \tan(90^\circ + 30^\circ)

    =−cos30∘.1−cos60∘−cot⁡30∘= -cos 30^\circ. \frac{1}{-cos 60^\circ} - \cot 30^\circ

    =32.21−3=0= \frac{\sqrt{3}}{2}.\frac{2}{1} - \sqrt{3} = 0

  24. Given, tan⁡(11π12)\tan\left(\frac{11\pi}{12}\right)

    =tan⁡(π−π12)=−tan⁡15∘= \tan\left(\pi - \frac{\pi}{12}\right) = -\tan15^\circ

    Using the value computed in 20 for tan⁡15∘\tan15^\circ we have 3−2\sqrt{3} - 2 as the answer.

  25. We know that tan(−θ)=−tan⁡θ,tan(-\theta) = -\tan\theta, thus

    tan⁡((−1)nπ4)=(−1)ntan⁡π4=(−1)n\tan \left((-1)^n\frac{\pi}{4}\right) = (-1)^n\tan\frac{\pi}{4} = (-1)^n

  26. Given, cos⁡18∘−sin⁡18∘=2sin⁡27∘\cos 18^\circ - \sin 18^\circ = \sqrt{2}\sin 27^\circ

    12cos⁡18∘−12sin⁡18∘=sin⁡27∘\frac{1}{\sqrt{2}}\cos 18^\circ - \frac{1}{\sqrt{2}}\sin18^\circ = \sin 27^\circ

    L.H.S. =sin⁡45∘cos⁡18∘−cos⁡45∘sin⁡18∘= \sin45^\circ\cos18^\circ - \cos45^\circ\sin18^\circ

    =sin⁡(45∘−18∘)=sin⁡27∘== \sin(45^\circ - 18^\circ) = \sin 27^\circ = R.H.S.

  27. L.H.S. =tan⁡70∘=tan⁡(50∘+20∘)=\tan 70^\circ = \tan(50^\circ + 20^\circ)

    =tan⁡50∘+tan⁡20∘1−tan⁡50∘tan⁡20∘= \frac{\tan 50^\circ + \tan 20^\circ}{1 - \tan50^\circ\tan20^\circ}

    tan⁡70∘−tan⁡70∘tan⁡50∘tan⁡20∘=tan⁡50∘+tan⁡20∘\tan70^\circ - \tan70^\circ\tan50^\circ\tan20^\circ = \tan 50^\circ + \tan 20^\circ

    tan⁡70∘=tan⁡70∘tan⁡50∘tan⁡20∘+tan⁡50∘+tan⁡20∘\tan70^\circ = \tan70^\circ\tan50^\circ\tan20^\circ + \tan 50^\circ + \tan 20^\circ

    =tan⁡(90∘−20∘)tan⁡50∘tan⁡20∘+tan⁡50∘+tan⁡20∘= \tan(90^\circ - 20^\circ)\tan50^\circ\tan20^\circ + \tan 50^\circ + \tan 20^\circ

    =cot⁡20∘tan⁡50∘tan⁡20∘+tan⁡50∘+tan⁡20∘= \cot20^\circ\tan50^\circ\tan20^\circ + \tan 50^\circ + \tan 20^\circ

    =tan⁡50∘+tan⁡50∘+tan⁡20∘=2tan⁡50∘+tan⁡20∘== \tan 50^\circ + \tan 50^\circ + \tan 20^\circ = 2\tan50^\circ + \tan20^\circ = R.H.S.

  28. L.H.S. =cos⁡(π4+x)cos⁡(π4−x)sin⁡(π4+x)sin⁡(π4−x)= \frac{\cos\left(\frac{\pi}{4} + x\right)\cos\left(\frac{\pi}{4} - x\right)}{\sin\left(\frac{\pi}{4} + x\right)\sin\left(\frac{\pi}{4} - x\right)}

    =cos⁡2π4−sin⁡2xsin⁡2π4−sin⁡2x=12−sin⁡2x12−sin⁡2x=1== \frac{\cos^2\frac{\pi}{4} - \sin^2x}{\sin^2\frac{\pi}{4} - \sin^2x} = \frac{\frac{1}{2} - \sin^2x}{\frac{1}{2} - \sin^2x} = 1 = R.H.S.

  29. L.H.S. =cos⁡(m+n)θ.cos⁡(m−n)θ−sin⁡(m+n)θsin⁡(m−n)θ= \cos(m + n)\theta.\cos(m - n)\theta - \sin(m + n)\theta\sin(m - n)\theta

    =cos⁡(m+n+m−n)θ=cos⁡2mθ== \cos(m + n + m - n)\theta = \cos2m\theta = R.H.S.

  30. L.H.S. =tan⁡(θ+ϕ)+tan⁡(θ−ϕ)1−tan⁡(θ+ϕ)tan⁡(θ−ϕ)= \frac{\tan(\theta + \phi) + \tan(\theta - \phi)}{1 - \tan(\theta + \phi)\tan(\theta - \phi)}

    =tan⁡(θ+ϕ+θ−ϕ)=tan⁡2θ== \tan(\theta + \phi + \theta - \phi) = \tan 2\theta = R.H.S.

  31. Given cos⁡9∘+sin⁡9∘=2sin⁡54∘\cos 9^\circ + \sin 9^\circ = \sqrt{2}\sin 54^\circ

    12cos⁡9∘+12sin⁡9∘=sin⁡54∘\frac{1}{\sqrt{2}}\cos9^\circ + \frac{1}{\sqrt{2}}\sin9^\circ = \sin54^\circ

    L.H.S. =sin⁡45∘cos⁡9∘+cos⁡45∘sin⁡9∘= \sin45^\circ\cos9^\circ + \cos45^\circ\sin9^\circ

    =sin⁡(45∘+9∘)=sin⁡54∘== \sin(45^\circ + 9^\circ) = \sin 54^\circ = R.H.S.

  32. L.H.S. =cos⁡20∘−sin⁡20∘cos⁡20∘+sin⁡20∘= \frac{\cos 20^\circ - \sin 20^\circ}{\cos 20^\circ + \sin 20^\circ}

    Dividing both numerator and denominaor with cos⁡20∘,\cos20^\circ, we get

    =1−tan⁡20∘1+tan⁡20∘=tan⁡45∘−tan⁡20∘1−tan⁡45∘tan⁡20∘ [∵tan⁡45∘=1]= \frac{1 - \tan20^\circ}{1 + \tan20^\circ} = \frac{\tan 45^\circ - \tan20^\circ}{1 - \tan45^\circ\tan20^\circ}~[\because \tan45^\circ = 1]

    =tan⁡(45∘−20∘)=tan⁡25∘== \tan(45^\circ - 20^\circ) = \tan 25^\circ = R.H.S.

  33. L.H.S. =tan⁡A+tan⁡Btan⁡A−tan⁡B= \frac{\tan A + \tan B}{\tan A - \tan B}

    =sin⁡Acos⁡A+sin⁡Bcos⁡Bsin⁡Acos⁡A−sin⁡Bcos⁡B= \frac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A} - \frac{\sin B}{\cos B}}

    =sin⁡Acos⁡B+sin⁡Bcos⁡Asin⁡Acos⁡B−sin⁡Bcos⁡A=sin⁡(A+B)sin⁡(A−B)== \frac{\sin A\cos B + \sin B\cos A}{\sin A\cos B - \sin B\cos A} = \frac{\sin(A + B)}{\sin (A - B)} = R.H.S.

  34. L.H.S. =1tan⁡3A−tan⁡A−1cot⁡3A−cot⁡A= \frac{1}{\tan 3A - \tan A} - \frac{1}{\cot 3A - \cot A}

    =1tan⁡3A−tan⁡A−11tan⁡3A−1tan⁡A= \frac{1}{\tan 3A - \tan A} - \frac{1}{\frac{1}{\tan 3A} - \frac{1}{\tan A}}

    =1tan⁡3A−tan⁡A−tan⁡Atan⁡3Atan⁡A−tan⁡3A= \frac{1}{\tan 3A - \tan A} - \frac{\tan A\tan 3A}{\tan A - \tan 3A}

    =1+tan⁡Atan⁡3Atan⁡3A−tan⁡A=1tan⁡(3A−A)=cot⁡2A== \frac{1 + \tan A \tan 3A}{\tan 3A - \tan A} = \frac{1}{\tan(3A - A)} = \cot 2A = R.H.S.

  35. This is similar to previous problema and can be solved likewise.

  36. L.H.S. =sin⁡3αsin⁡α+cos⁡3αcosα= \frac{\sin 3\alpha}{\sin\alpha} + \frac{\cos 3\alpha}{cos\alpha}

    =sin⁡3αcos⁡α+cos⁡3αsin⁡αsin⁡αcos⁡α= \frac{\sin3\alpha\cos\alpha + \cos3\alpha\sin\alpha}{\sin\alpha\cos\alpha}

    =sin⁡(3α+α)sin⁡αcos⁡α= \frac{\sin(3\alpha + \alpha)}{\sin\alpha\cos\alpha}

    =22sin⁡4αsin⁡2α [∵sin⁡αcos⁡α=12sin⁡2α]=2\frac{2\sin 4\alpha}{\sin2\alpha}~[\because \sin\alpha\cos\alpha = \frac{1}{2}\sin2\alpha]

    =22sin⁡2αcos⁡2αsin⁡2α=4cos⁡2α== 2\frac{2\sin2\alpha\cos2\alpha}{\sin2\alpha} = 4\cos2\alpha = R.H.S.

  37. L.H.S. =tan⁡(π4+A)−tan⁡(π4−A)tan⁡(π4+A)+tan⁡(π4−A)= \frac{\tan\left(\frac{\pi}{4} + A \right) - \tan\left(\frac{\pi}{4} - A\right)}{\tan\left(\frac{\pi}{4} + A\right) + \tan\left(\frac{\pi}{4} - A\right)}

    =1+tan⁡A1−tan⁡A−1−tan⁡A1+tan⁡A1+tan⁡A1−tan⁡A+1−tan⁡A1+tan⁡A= \frac{\frac{1 + \tan A}{1 - \tan A} - \frac{1 - \tan A}{1 + \tan A}}{\frac{1 + \tan A}{1 - \tan A} + \frac{1 - \tan A}{1 + \tan A}}

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

    =4tan⁡A2+2tan⁡2A=2tan⁡Asec⁡2A=2sin⁡Acos⁡A=sin⁡2A= \frac{4\tan A}{2 + 2\tan^2A} = \frac{2\tan A}{\sec^2A} = 2\sin A\cos A = \sin 2A

  38. Given, tan⁡40∘+2tan⁡10∘=tan⁡50∘\tan 40^\circ + 2 \tan 10^\circ = \tan 50^\circ

    R.H.S. =tan⁡50∘=tan⁡(40∘+10∘)= \tan 50^\circ = \tan(40^\circ + 10^\circ)

    =tan⁡40∘+tan⁡10∘1−tan⁡40∘tan⁡10∘= \frac{\tan 40^\circ + \tan 10^\circ}{1 - \tan 40^\circ\tan 10^\circ}

    tan⁡50∘−tan⁡50∘tan⁡40∘tan⁡10∘=tan⁡40∘+tan⁡10∘\tan50^\circ - \tan50^\circ\tan40^\circ\tan10^\circ = \tan 40^\circ + \tan 10^\circ

    tan⁡50∘−cot⁡40∘tan⁡40∘tan⁡10∘=tan40∘+tan⁡10∘\tan50^\circ - \cot40^\circ\tan40^\circ\tan10^\circ = tan40^\circ + \tan10^\circ

    tan⁡50∘=tan⁡40∘+2tan⁡10∘\tan50^\circ = \tan 40^\circ + 2 \tan 10^\circ

  39. R.H.S. =tan⁡(α+β)tan⁡(α−β)= \tan(\alpha + \beta)\tan(\alpha - \beta)

    =sin⁡(α+β)cos⁡(α+β)sin⁡(α−β)cos⁡(α−β)= \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)}\frac{\sin(\alpha -\beta)}{\cos(\alpha - \beta)}

    =sin⁡αcos⁡β+cos⁡αsin⁡βcos⁡αcos⁡β−sin⁡αsin⁡βsin⁡αcos⁡β−cos⁡αsin⁡βcos⁡αcos⁡β+sin⁡αsin⁡β= \frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\cos\alpha\cos\beta - \sin\alpha\sin\beta}\frac{\sin\alpha\cos\beta - \cos\alpha\sin\beta}{\cos\alpha\cos\beta + \sin\alpha\sin\beta}

    =sin⁡2αcos⁡2β−sin⁡2βcos⁡2αcos⁡2αcos⁡2β−sin⁡2αsin⁡2β= \frac{\sin^2\alpha\cos^2\beta - \sin^2\beta\cos^2\alpha}{\cos^2\alpha\cos^2\beta - \sin^2\alpha\sin^2\beta}

    =sin⁡2α(1−sin⁡2β)−sin⁡2β(1−sin⁡2α)cos⁡2α(1−sin⁡2β)−sin⁡2β(1−cos⁡2α)= \frac{\sin^2\alpha(1 - \sin^2\beta) - \sin^2\beta(1 - \sin^2\alpha)}{\cos^2\alpha(1 - \sin^2\beta) - \sin^2\beta(1 - \cos^2\alpha)}

    =sin⁡2α−sin⁡2βcos⁡2α−sin⁡2β== \frac{\sin^2\alpha - \sin^2\beta}{\cos^2\alpha - \sin^2\beta} = R.H.S.

  40. L.H.S. =tan⁡2α−tan⁡2β=sin⁡2αcos⁡2α−sin⁡2βcos⁡2β= \tan^2\alpha -\tan^2\beta = \frac{\sin^2\alpha}{\cos^2\alpha} - \frac{\sin^2\beta}{\cos^2\beta}

    =sin⁡2αcos⁡2β−sin⁡2βcos⁡2αcos⁡2αcos⁡2β= \frac{\sin^2\alpha\cos^2\beta - \sin^2\beta\cos^2\alpha}{\cos^2\alpha\cos^2\beta}

    =(sin⁡αcos⁡β+sin⁡βsin⁡α)(sin⁡αcos⁡β−sin⁡βsin⁡α)cos⁡2αcos⁡2β= \frac{(\sin\alpha\cos\beta + \sin\beta\sin\alpha)(\sin\alpha\cos\beta - \sin\beta\sin\alpha)}{\cos^2\alpha\cos^2\beta}

    =sin⁡(α+β)sin⁡(α−β)cos⁡2αcos⁡2β== \frac{\sin(\alpha + \beta)\sin(\alpha - \beta)}{\cos^2\alpha\cos^2\beta} = R.H.S.

  41. L.H.S. =tan⁡[(2n+1)π+θ]+tan⁡[(2n+1)π−θ]= \tan[(2n + 1)\pi + \theta] + \tan[(2n + 1)\pi - \theta]

    =tan⁡(π+θ)+tan⁡(π−θ) [∵tan⁡2nπ=0]= \tan(\pi + \theta) + \tan(\pi - \theta)~[\because \tan 2n\pi = 0]

    =tan⁡θ−tan⁡θ=0== \tan\theta - \tan\theta = 0 = R.H.S.

  42. L.H.S. =tan⁡(π4+θ)tan⁡(3π4+θ)+1= \tan\left(\frac{\pi}{4} + \theta\right)\tan\left(\frac{3\pi}{4} + \theta\right) + 1

    =tan⁡(π4+θ)tan⁡[π−(π4−θ)]+1= \tan\left(\frac{\pi}{4} + \theta\right)\tan\left[\pi - \left(\frac{\pi}{4} - \theta\right)\right] + 1

    =−tan⁡(π4+θ)tan⁡(π4−θ)+1= -\tan\left(\frac{\pi}{4} + \theta\right)\tan\left(\frac{\pi}{4} - \theta\right) + 1

    =−tan⁡(π4+θ)tan⁡[(π2−pi4−θ)]+1= -\tan\left(\frac{\pi}{4} + \theta\right)\tan\left[\left(\frac{\pi}{2} - \frac{pi}{4} - \theta\right)\right] + 1

    =−tan⁡(π4+θ)cot⁡(π4+θ)+1=−1+1=0== -\tan\left(\frac{\pi}{4} + \theta\right)\cot\left(\frac{\pi}{4} + \theta\right) + 1 = -1 + 1 = 0 = R.H.S.

  43. R.H.S. =1−pq(1+p2)(1+q2)= \frac{1 - pq}{\sqrt{(1 + p^2)(1 + q^2)}}

    Substituting for pp and q,q, we get

    =1−tan⁡αtan⁡β(1+tan⁡2α)(1+tan⁡2β)= \frac{1 - \tan\alpha\tan\beta}{\sqrt{(1 + \tan^2\alpha)(1 + \tan^2\beta)}}

    =cos⁡αcos⁡β−sin⁡αsin⁡βcos⁡αcos⁡βsec⁡αsec⁡β= \frac{\frac{\cos\alpha\cos\beta - \sin\alpha\sin\beta}{\cos\alpha\cos\beta}}{\sqrt{\sec\alpha\sec\beta}}

    =cos⁡(α+β)== \cos(\alpha + \beta) = R.H.S.

  44. Given, tan⁡β=2sin⁡αsin⁡γsin⁡(α+γ)\tan \beta = \frac{2\sin\alpha\sin\gamma}{\sin(\alpha + \gamma)}

    Inverting, we get

    2cot⁡β=sin⁡(α+γ)sin⁡αsin⁡γ=sin⁡αcos⁡γ+sin⁡γcos⁡αsin⁡αsin⁡γ2\cot\beta = \frac{\sin(\alpha + \gamma)}{\sin\alpha\sin\gamma} = \frac{\sin\alpha\cos\gamma + \sin\gamma\cos\alpha}{\sin\alpha\sin\gamma}

    =cot⁡α+cot⁡γ= \cot \alpha + \cot \gamma

    Thus, cot⁡α,cot⁡β,cot⁡γ\cot\alpha, \cot\beta, \cot\gamma are in A.P.

  45. tan⁡(θ+α−(θ−α))=tan⁡2α=tan⁡(θ+α)−tan⁡(θ−α)1+tan⁡(θ+α)tan⁡(θ−α)\tan(\theta + \alpha - (\theta - \alpha)) = \tan2\alpha = \frac{\tan(\theta + \alpha) - \tan(\theta - \alpha)}{1 + \tan(\theta + \alpha)\tan(\theta - \alpha)}

    =b−a1+ab= \frac{b - a}{1 + ab}

  46. tan⁡γ=tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β\tan\gamma = \tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}

    a−b=cot⁡α+cot⁡β−tan⁡α−tan⁡βa - b = \cot\alpha + \cot\beta - \tan\alpha - \tan\beta

    =sin⁡(α+β)sin⁡αsin⁡β−sin⁡(α+β)cos⁡αβ= \frac{\sin(\alpha + \beta)}{\sin\alpha\sin\beta} - \frac{\sin(\alpha + \beta)}{\cos\alpha\beta}

    =sin⁡(α+β)(cos⁡αcos⁡β−sin⁡αsin⁡βsin⁡αsin⁡βcos⁡αcos⁡β)= \sin(\alpha + \beta)\left(\frac{\cos\alpha\cos\beta - \sin\alpha\sin\beta}{\sin\alpha\sin\beta\cos\alpha\cos\beta}\right)

    =sin⁡(α+β)cos⁡(α+β)sin⁡αsin⁡βcos⁡αcos⁡β= \frac{\sin(\alpha + \beta)\cos(\alpha + \beta)}{\sin\alpha\sin\beta\cos\alpha\cos\beta}

    ab=(tan⁡α+tan⁡β)(cot⁡α+cot⁡β)ab = (\tan\alpha + \tan\beta)(\cot\alpha + \cot\beta)

    =sin⁡2(α+β)sin⁡αsin⁡βcos⁡αcos⁡β= \frac{\sin^2(\alpha + \beta)}{\sin\alpha\sin\beta\cos\alpha\cos\beta}

    aba−b=tan⁡(α+β)=tan⁡γ\frac{ab}{a - b} = \tan(\alpha + \beta) = \tan\gamma

  47. Given, A+B=45∘∴tan⁡(A+B)=1A + B = 45^\circ \therefore \tan(A + B) = 1

    tan⁡A+tan⁡B1−tan⁡Atan⁡B=1\frac{\tan A + \tan B}{1 - \tan A\tan B} = 1

    1+tan⁡A+tan⁡B+tan⁡Atan⁡B=21 + \tan A + \tan B + \tan A \tan B = 2

    (1+tan⁡A)(1+tan⁡B)=2(1 + \tan A)(1 + \tan B) = 2

  48. Given, sin⁡αsin⁡β−cos⁡αcos⁡β+1=0\sin\alpha\sin\beta - \cos\alpha\cos\beta + 1 = 0

    ⇒cos⁡αcos⁡β−sin⁡αsin⁡β=1\Rightarrow \cos\alpha\cos\beta - \sin\alpha\sin\beta = 1

    ⇒cos⁡(α+β)=1\Rightarrow \cos(\alpha + \beta) = 1

    ⇒sin⁡(α+β)=0\Rightarrow \sin(\alpha + \beta) = 0

    1+cot⁡αtan⁡β=1+cos⁡αsin⁡βsin⁡αcos⁡β1 + \cot\alpha\tan\beta = 1 + \frac{\cos\alpha\sin\beta}{\sin\alpha\cos\beta}

    =sin⁡αcos⁡β+cos⁡αsin⁡βsin⁡αcos⁡β= \frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\sin\alpha\cos\beta}

    =sin⁡(α+β)sin⁡αcos⁡β=0sin⁡αcos⁡β=0= \frac{\sin(\alpha + \beta)}{\sin\alpha\cos\beta} = \frac{0}{\sin\alpha\cos\beta} = 0

  49. tan⁡β=nsin⁡αcos⁡α1−nsin⁡2α=nsin⁡αcos⁡αcos⁡2α1cos⁡2α−nsin⁡2αcos⁡2α\tan\beta = \frac{n\sin\alpha\cos\alpha}{1 - n\sin^2\alpha} = \frac{\frac{n\sin\alpha\cos\alpha}{\cos^2\alpha}}{\frac{1}{\cos^2\alpha} - n\frac{\sin^2\alpha}{\cos^2\alpha}}

    =ntan⁡αsec⁡2α−ntan⁡2α=ntan⁡α1+(1−n)tan⁡2α= \frac{n\tan\alpha}{\sec^2\alpha - n\tan^2\alpha} = \frac{n\tan\alpha}{1 + (1 - n)\tan^2\alpha}

    Now, tan⁡(α−β)=tan⁡α−ntan⁡α1+(1−n)tan⁡2α1−tan⁡αntan⁡α1+(1−n)tan⁡2α\tan(\alpha - \beta) = \frac{\tan\alpha - \frac{n\tan\alpha}{1 + (1 - n)\tan^2\alpha}}{1 - \tan\alpha\frac{n\tan\alpha}{1 + (1 - n)\tan^2\alpha}}

    =tan⁡α+(1−n)tan⁡3α−ntan⁡α1+(1−n)tan⁡2α+ntan⁡2α= \frac{\tan\alpha + (1 - n)\tan^3\alpha - n\tan\alpha}{1 + (1 - n)\tan^2\alpha + n\tan^2\alpha}

    =(1−n)tan⁡α+(1−n)tan3α1+tan⁡2α= \frac{(1 - n)\tan\alpha + (1 - n)tan^3\alpha}{1 + \tan^2\alpha}

    =(1−n)tan⁡α(1+tan⁡2α)1+tan⁡2α=(1−n)tan⁡α= \frac{(1 - n)\tan\alpha(1 + \tan^2\alpha)}{ 1 + \tan^2\alpha} = (1 - n)\tan\alpha

  50. Given, cos⁡(β−γ)+cos⁡(γ−α)+cos⁡(α−β)=−32\cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta) = -\frac{3}{2}

    3+2cos⁡(β−γ)+2cos⁡(γ−α)+2cos⁡(α−β)=03 + 2\cos(\beta - \gamma) + 2\cos(\gamma - \alpha) + 2\cos(\alpha - \beta) = 0

    3+2(cos⁡βcos⁡γ+sin⁡βsin⁡γ)+2(cos⁡γcos⁡α+sin⁡γsin⁡α)+2(cos⁡αcos⁡β+sin⁡αsin⁡β)=03 + 2(\cos\beta\cos\gamma + \sin\beta\sin\gamma) + 2(\cos\gamma\cos\alpha + \sin\gamma\sin\alpha) + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) = 0

    (cos⁡2α+sin⁡2α)+(cos⁡2β+sin⁡2β)+(cos⁡2γ+sin⁡2γ)+2(cos⁡βcos⁡γ+sin⁡βsin⁡γ)+2(cos⁡γcos⁡α+sin⁡γsin⁡α)+2(cos⁡αcos⁡β+sin⁡αsin⁡β)=0(\cos^2\alpha + \sin^2\alpha) + (\cos^2\beta + \sin^2\beta) + (\cos^2\gamma + \sin^2\gamma) + 2(\cos\beta\cos\gamma + \sin\beta\sin\gamma) + 2(\cos\gamma\cos\alpha + \sin\gamma\sin\alpha) + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) = 0

    (cos⁡α+cos⁡β+cos⁡γ)2+(sin⁡α+sin⁡β+sin⁡γ2)=0(\cos\alpha + \cos\beta + \cos\gamma)^2 + (\sin\alpha + \sin\beta + \sin\gamma^2) = 0

    cos⁡α+cos⁡β+cos⁡γ=sin⁡α+sin⁡β+sin⁡γ=0\cos\alpha + \cos\beta + \cos\gamma = \sin\alpha + \sin\beta + \sin\gamma = 0

  51. tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}

    =mm+1+12m+11−mm+112m+1= \frac{\frac{m}{m + 1} + \frac{1}{2m + 1}}{1 - \frac{m}{m + 1}\frac{1}{2m + 1}}

    =2m2+m+m+12m2+3m+1−n=1= \frac{2m^2 + m + m + 1}{2m^2 + 3m + 1 - n} = 1

    Thus, α+beta=π4\alpha + beta = \frac{\pi}{4}

  52. Given (cot⁡A−1)(cot⁡B−1)=2(\cot A - 1)(\cot B - 1) = 2

    cot⁡Acot⁡B−1−cot⁡A−cot⁡B=0\cot A\cot B - 1 - \cot A - \cot B = 0

    cot⁡Acot⁡B−1=cot⁡A+cot⁡B⇒cot⁡Acot⁡B−1cot⁡A+cot⁡B=1\cot A\cot B - 1 = \cot A + \cot B \Rightarrow \frac{\cot A\cot B - 1}{\cot A + \cot B} = 1

    cot⁡(A+B)=cot⁡45∘\cot(A + B) = \cot 45^\circ

    Thus, A+B=45∘A + B = 45^\circ

    which we have proved in reverse.

  53. Given, tan⁡α−tan⁡β=x\tan\alpha - \tan\beta = x and cot⁡β−cot⁡α=y,\cot\beta - \cot\alpha = y, we have to prove that cot⁡(α−β)=x+yxy\cot(\alpha - \beta) = \frac{x + y}{xy}

    Let cot⁡(α−β)=x+yxy=tan⁡α−tan⁡β+cot⁡β−cot⁡α(tan⁡α−tan⁡β)(cot⁡β−cot⁡α)\cot(\alpha - \beta) = \frac{x + y}{xy} = \frac{\tan\alpha - \tan\beta + \cot\beta - \cot\alpha}{(\tan\alpha - \tan\beta)(\cot\beta - \cot\alpha)}

    tan⁡(α−β)=(tan⁡α−tan⁡β)(cot⁡β−cot⁡α)tan⁡α−tan⁡β+cot⁡β−cot⁡α\tan(\alpha - \beta) = \frac{(\tan\alpha - \tan\beta)(\cot\beta - \cot\alpha)}{\tan\alpha - \tan\beta + \cot\beta - \cot\alpha}

    =tan⁡α−tan⁡β1+tan⁡α−tan⁡βcot⁡β−cot⁡α= \frac{\tan\alpha - \tan\beta}{1 + \frac{\tan\alpha - \tan\beta}{\cot\beta - \cot \alpha}}

    =tan⁡α−tan⁡β1+sin⁡(α−β)cos⁡αcos⁡βsin⁡(α−β)cos⁡αcos⁡β= \frac{\tan\alpha - \tan\beta}{ 1 + \frac{\frac{\sin(\alpha - \beta)}{\cos\alpha\cos\beta}}{\frac{\sin(\alpha - \beta)}{\cos\alpha\cos\beta}}}

    =tan⁡α−tan⁡β1+tan⁡αtan⁡β=tan⁡(α−β)= \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta} = \tan(\alpha - \beta)

    Hence proved.

  54. Given α+β+γ=90∘=π2\alpha + \beta + \gamma = 90^\circ = \frac{\pi}{2}

    cot⁡α=cot⁡(π2−(β+γ))=tan⁡(β+γ)\cot \alpha = \cot\left(\frac{\pi}{2} - (\beta + \gamma)\right) = \tan(\beta + \gamma)

    =tan⁡β+tan⁡γ1−tan⁡βtan⁡γ= \frac{\tan\beta + \tan\gamma}{1 - \tan\beta\tan\gamma}

  55. We have to prove that cot⁡β=2tan⁡(α−β)\cot \beta = 2\tan(\alpha - \beta)

    1tan⁡β=2tan⁡α−tan⁡β1+tan⁡αtan⁡β\frac{1}{\tan\beta} = 2\frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}

    1+tan⁡αtan⁡β=2tan⁡αtan⁡β−2tan⁡2β1 + \tan\alpha\tan\beta = 2\tan\alpha\tan\beta - 2\tan^2\beta

    Dividing both sides by tan⁡β,\tan\beta, we get

    cot⁡β+tan⁡α=2tan⁡α−2tan⁡β\cot \beta + \tan\alpha = 2\tan\alpha - 2\tan\beta

    cot⁡β+2tan⁡β==tan⁡α\cot\beta + 2\tan\beta = =\tan\alpha

    Hence proved.

  56. sin⁡A=ac,sin⁡B=bc,cos⁡A=bc,cos⁡B=ac\sin A = \frac{a}{c}, \sin B = \frac{b}{c}, \cos A = \frac{b}{c}, \cos B = \frac{a}{c}

    cosec⁡(A−B)=1sin⁡(A−B)=1sin⁡Acos⁡B−cos⁡Asin⁡B=1a2c2−b2c2\cosec(A - B) = \frac{1}{\sin(A - B)} = \frac{1}{\sin A\cos B - \cos A\sin B} = \frac{1}{\frac{a^2}{c^2} - \frac{b^2}{c^2}}

    =c2a2−b2=a2+b2a2−b2= \frac{c^2}{a^2 - b^2} = \frac{a^2 + b^2}{a^2 - b^2}

    sec⁡(A−B)=1cos⁡(A−B)=1cos⁡Acos⁡B+sin⁡Asin⁡B=c22ab\sec(A - B) = \frac{1}{\cos(A - B)} = \frac{1}{\cos A\cos B + \sin A\sin B} = \frac{c^2}{2ab}

  57. We have to prove that A+B=CA + B = C i.e. tan⁡(A+B)=tan⁡C\tan(A + B) = \tan C

    tan⁡A+tan⁡B1−tan⁡Atan⁡B=tan⁡C\frac{\tan A + \tan B}{1 - \tan A\tan B} = \tan C

    1ac+ac1−1acac=ca3\frac{\frac{1}{\sqrt{ac}} + \sqrt{\frac{a}{c}}}{1 - \frac{1}{\sqrt{ac}}\sqrt{\frac{a}{c}}} = \sqrt{\frac{c}{a^3}}

    =1ac+ac1−1c= \frac{\frac{1}{\sqrt{ac}} + \sqrt{\frac{a}{c}}}{1 - \frac{1}{c}}

    1+aac.cc−1=ca3\frac{1 + a}{\sqrt{ac}}.\frac{c}{c - 1} = \sqrt{\frac{c}{a^3}}

    ac+cc−1=ca\frac{ac + c}{c - 1} = \frac{c}{a}

    a2c+ac=c2−ca^2c + ac = c^2 - c

    a2+a+1=ca^2 + a + 1 = c which is given, hence proved.

  58. Given tan⁡(A−B)tan⁡A+sin⁡2Csin⁡2A=1\frac{\tan(A - B)}{\tan A} + \frac{\sin^2C}{\sin^2A} = 1

    sin⁡2Csin⁡2A=1−tan⁡(A−B)tan⁡A\frac{\sin^2C}{\sin^2A} = 1 - \frac{\tan(A - B)}{\tan A}

    =1−sin⁡(A−B)cos⁡Asin⁡Acos⁡(A−B)=sin⁡(A−A+B)sin⁡Acos⁡(A−B)= 1 - \frac{\sin(A - B)\cos A}{\sin A\cos(A - B)} = \frac{\sin (A - A + B)}{\sin A\cos(A - B)}

    sin⁡2C=sin⁡Asin⁡Bcos⁡(A−B)\sin^2C = \frac{\sin A\sin B}{\cos(A - B)}

    cosec⁡2C=cos⁡(A−B)sin⁡Asin⁡B=1+cot⁡Acot⁡B=cot⁡2C\cosec^2C = \frac{\cos(A - B)}{\sin A\sin B} = 1 + \cot A\cot B = \cot^2C

    ⇒tan⁡Atan⁡B=tan⁡2C\Rightarrow \tan A\tan B= \tan^2 C

  59. Given, sin⁡αsin⁡β−cos⁡αcos⁡β=1\sin\alpha\sin\beta - \cos\alpha\cos\beta = 1

    cos⁡(α+β)=−1⇒α+β=(2n+1)π\cos(\alpha + \beta) = -1 \Rightarrow \alpha + \beta = (2n + 1)\pi

    tan⁡(α+β)=0⇒tan⁡α+tan⁡β=0\tan(\alpha + \beta) = 0 \Rightarrow \tan \alpha + \tan \beta = 0

  60. Given, sin⁡θ=3sin⁡(θ+2α)\sin\theta = 3\sin(\theta + 2\alpha)

    sin⁡(θ+α−α)=3sin⁡(θ+α+α)\sin(\theta + \alpha - \alpha) = 3\sin(\theta + \alpha + \alpha)

    sin⁡(θ+α)cos⁡α−sin⁡αcos⁡(θ+α)=3sin⁡(θ+α)cos⁡α+3cos⁡(θ+α)sin⁡α\sin(\theta + \alpha)\cos\alpha - \sin\alpha\cos(\theta + \alpha) = 3\sin(\theta + \alpha)\cos\alpha + 3\cos(\theta + \alpha)\sin\alpha

    2sin⁡(θ+α)cos⁡α+4sin⁡αcos⁡(θ+α)=02\sin(\theta + \alpha)\cos\alpha + 4\sin\alpha\cos(\theta + \alpha) = 0

    Dividingboth sides with 2cos⁡(θ+α)cos⁡α,2\cos(\theta + \alpha)\cos\alpha, we get

    tan⁡(θ+α)+2tan⁡α=0\tan(\theta + \alpha) + 2\tan\alpha = 0

  61. Given, 3tan⁡θtan⁡ϕ=1⇒cot⁡θcot⁡ϕ=33\tan\theta\tan\phi = 1 \Rightarrow \cot\theta\cot\phi = 3

    cos⁡θcos⁡ϕsin⁡θsin⁡ϕ=3\frac{\cos\theta\cos\phi}{\sin\theta\sin\phi} = 3

    Applying componendo and dividendo

    cos⁡θcos⁡ϕ+sin⁡θsin⁡ϕcos⁡θcos⁡ϕ−sin⁡θsin⁡ϕ=3+13−1\frac{\cos\theta\cos\phi + \sin\theta\sin\phi}{\cos\theta\cos\phi - \sin\theta\sin\phi} = \frac{3 + 1}{3 - 1}

    cos⁡(θ−ϕ)=2cos⁡(θ+ϕ)\cos(\theta - \phi) = 2\cos(\theta + \phi)

  62. Let z=cos⁡θ+sin⁡θ=2(12cos⁡θ+12sin⁡θ)z = \cos\theta + \sin\theta = \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos\theta + \frac{1}{\sqrt{2}}\sin\theta\right)

    =2cos⁡(θ−π4)= \sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right)

    =2cos⁡55∘= \sqrt{2}\cos 55^\circ which has positive sign.

  63. Let z=5cos⁡θ+3cos⁡(θ+π3)+3z = 5\cos\theta + 3\cos\left(\theta + \frac{\pi}{3}\right) + 3

    z=5cos⁡θ+32cos⁡θ−332sin⁡θ+3z = 5\cos\theta + \frac{3}{2}\cos\theta - \frac{3\sqrt{3}}{2}\sin\theta + 3

    =132cos⁡θ−332sin⁡θ+3= \frac{13}{2}\cos\theta - \frac{3\sqrt{3}}{2}\sin\theta + 3

    =7(1314cos⁡θ−3314sin⁡θ)+3= 7\left(\frac{13}{14}\cos\theta - \frac{3\sqrt{3}}{14}\sin\theta\right) + 3

    Let cos⁡α=1314\cos\alpha = \frac{13}{14} then sin⁡α=3314\sin\alpha = \frac{3\sqrt{3}}{14}

    y=7(cos⁡αcos⁡θ−sin⁡αsin⁡θ)+3y = 7(\cos\alpha\cos\theta - \sin\alpha\sin\theta) + 3

    y=7cos⁡(θ+α)+3y = 7\cos(\theta + \alpha) + 3

    Now maximum and minimum values of cos⁡(θ+α)\cos(\theta + \alpha) are 11 and −1.-1. Thus, value of yy will lie between 44 and 10.10.

  64. Given, mtan⁡(θ−30∘)=ntan⁡(θ+120∘)m\tan(\theta - 30^\circ) = n\tan(\theta + 120^\circ)

    tan⁡(θ−30∘)tan⁡(θ+120∘)=nm\frac{\tan(\theta - 30^\circ)}{\tan(\theta + 120^\circ)} = \frac{n}{m}

    sin⁡(θ−30∘)cos⁡(θ)+120∘cos⁡(θ−30∘)sin⁡(θ)+120∘=nm\frac{\sin(\theta - 30^\circ)\cos(\theta) + 120^\circ}{\cos(\theta - 30^\circ)\sin(\theta) + 120^\circ} = \frac{n}{m}

    Applying componendo and dividendo

    sin⁡[(θ+120∘)+(θ−30∘)]sin⁡[(θ+120∘)−(θ−30∘)=m+nm−n\frac{\sin[(\theta + 120^\circ) + (\theta - 30^\circ)]}{\sin[(\theta + 120^\circ) - (\theta - 30^\circ)} = \frac{m + n}{m - n}

    sin⁡(2θ+90∘)sin⁡150∘=m+nm−n\frac{\sin(2\theta + 90^\circ)}{\sin150^\circ} = \frac{m + n}{m - n}

    cos⁡2θ=m+n2(m−n)\cos2\theta = \frac{m + n}{2(m - n)}

  65. Given, tan⁡αtan⁡β=xy\frac{\tan\alpha}{\tan\beta} = \frac{x}{y}

    Applying componendo and dividendo

    tan⁡α+tan⁡βtan⁡α−tan⁡β=x+yx−y\frac{\tan\alpha + \tan\beta}{\tan\alpha - \tan\beta} = \frac{x + y}{x - y}

    sin⁡(α+β)sin⁡(α−β)=x+yx−y\frac{\sin(\alpha + \beta)}{\sin(\alpha - \beta)} = \frac{x + y}{x - y}

    sin⁡(α−β)=x−yx+ysin⁡θ\sin(\alpha - \beta) = \frac{x - y}{x + y}\sin\theta

  66. We have to find the maximum and minimul values of 7cos⁡θ+24sin⁡θ=y7\cos\theta + 24\sin\theta = y (let)

    =25(725cos⁡θ+2425sin⁡θ)= 25\left(\frac{7}{25}\cos\theta + \frac{24}{25}\sin\theta\right)

    If cos⁡α=725\cos\alpha = \frac{7}{25} then sin⁡α=2475\sin\alpha = \frac{24}{75}

    y=25cos⁡(θ−α)y = 25\cos(\theta - \alpha)

    Thus, maximum and minimum values of yy are 2525 and −25.-25.

  67. Given expression is sin⁡100∘−sin⁡10∘=cos⁡10∘−sin⁡10∘=y\sin100^\circ - \sin10^\circ = \cos10^\circ - \sin10^\circ = y (let)

    y=2(12cos⁡10−12sin⁡10∘)y = \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos^10 - \frac{1}{\sqrt{2}\sin10^\circ}\right)

    =2cos⁡(45∘+10∘)= \sqrt{2}\cos(45^\circ + 10^\circ)

    Thus, the sign is positive.