13. Multiple and Submultiple Angles Solutions#

  1. Let us solve these one by one.

    1. Given, cos⁡A=35\cos A = \frac{3}{5}

      ⇒sin⁡A=1−cos⁡2A=1−925=1625=45\Rightarrow \sin A = \sqrt{1 - \cos^2A} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}

      sin⁡2A=2sin⁡Acos⁡A=2.45.35=2425\sin 2A = 2\sin A\cos A = 2.\frac{4}{5}.\frac{3}{5} = \frac{24}{25}

    2. Given, sin⁡A=1213\sin A = \frac{12}{13}

      ⇒cos⁡A=1−sin⁡2A=1−144169=25169=513\Rightarrow \cos A = \sqrt{1 - \sin^2A} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}

      sin⁡2A=2sin⁡Acos⁡A=2.1213.513=120169\sin 2A = 2\sin A\cos A = 2.\frac{12}{13}.\frac{5}{13} = \frac{120}{169}

    3. Given, tan⁡A=1663=perpendicularbase\tan A = \frac{16}{63} = \frac{\text{perpendicular}}{\text{base}}

      hypotenuse=p2+b2=162+632=65\text{hypotenuse} = \sqrt{p^2 + b^2} = \sqrt{16^2 + 63^2} = 65

      sin⁡A=1665,cos⁡A=6365\sin A = \frac{16}{65}, \cos A = \frac{63}{65}

      sin⁡2A=2sin⁡Acos⁡A=2.1665.6365=20164225\sin 2A = 2\sin A\cos A = 2.\frac{16}{65}.\frac{63}{65} = \frac{2016}{4225}

  2. Let us solve these one by one.

    1. Given, cos⁡A=1517\cos A = \frac{15}{17}

      ⇒sin⁡A=1−cos⁡2A\Rightarrow \sin A = \sqrt{1 - \cos^2A} =1−225289=64289=817= \sqrt{1 - \frac{225}{289}} = \sqrt{\frac{64}{289}} = \frac{8}{17}

      cos⁡2A=cos⁡2A−sin⁡2A=225−64289=161289\cos 2A = \cos^2A - \sin^2A = \frac{225 - 64}{289} = \frac{161}{289}

    2. Given, sin⁡A=45\sin A = \frac{4}{5}

      ⇒cos⁡A=1−sin⁡2A\Rightarrow \cos A = \sqrt{1 - \sin^2A} =1−1625=35= \sqrt{1 - \frac{16}{25}} = \frac{3}{5}

      cos⁡2A=cos⁡2A−sin⁡2A=9−1625=−725\cos2A = \cos^2A - \sin^2A = \frac{9 - 16}{25} = -\frac{7}{25}

    3. Give, tan⁡A=512=perpendicularbase\tan A = \frac{5}{12} = \frac{\text{perpendicular}}{\text{base}}

      hypotenuse=p2+b2=25+144=13\text{hypotenuse} = \sqrt{p^2 + b^2} = \sqrt{25 + 144} = 13

      sin⁡A=513,cos⁡A=1213\sin A = \frac{5}{13}, \cos A = \frac{12}{13}

      cos⁡2A=cos⁡2A−sin⁡2A=119169\cos^2A = \cos^2A - \sin^2A = \frac{119}{169}

  3. Given, tan⁡A=ba,\tan A = \frac{b}{a}, thus hypotenuse=b2+a2\text{hypotenuse} = \sqrt{b^2 + a^2}

    acos⁡2A+bsin⁡2A=a(cos⁡2A−sin⁡2A)+2bsin⁡Acos⁡Aa\cos 2A+ b\sin 2A = a(\cos^2A - \sin^2A) + 2b\sin A\cos A

    =a(a2a2+b2−b2a2+b2)+2b.aba2+b2= a\left(\frac{a^2}{a^2 + b^2} - \frac{b^2}{a^2 + b^2}\right) + 2b.\frac{ab}{a^2 + b^2}

    =a(a2−b2+2b2a2+b2)=a= a\left(\frac{a^2 - b^2 + 2b^2}{a^2 + b^2}\right) = a

  4. We have to prove that sin⁡2A1+cos⁡2A=tan⁡A\frac{\sin 2A}{1 + \cos 2A} = \tan A

    L.H.S. =sin⁡2A1+cos⁡2A=2sin⁡Acos⁡A1+cos⁡2A−sin⁡2A= \frac{\sin 2A}{1 + \cos 2A} = \frac{2\sin A\cos A}{1 + \cos^2A - \sin^2A}

    =2sin⁡Acos⁡A2cos⁡2A[∵1−sin⁡2A=cos⁡2A]= \frac{2\sin A\cos A}{2\cos^2A}[\because 1 - \sin^2A = \cos^2A]

    =tan⁡A== \tan A = R.H.S.

  5. We have to prove that sin⁡2A1−cos⁡2A=cot⁡A\frac{\sin 2A}{1 - \cos 2A} = \cot A

    L.H.S. =sin⁡2A1−cos⁡2A=2sin⁡Acos⁡A1−(cos⁡2A−sin⁡2A)= \frac{\sin 2A}{1 - \cos 2A} = \frac{2\sin A\cos A}{1 -(\cos^2A - \sin^2A)}

    =2sin⁡Acos⁡A2sin⁡2A=cot⁡A== \frac{2\sin A\cos A}{2\sin^2A} = \cot A = R.H.S.

  6. We have to prove that 1−cos⁡2A1+cos⁡2A=tan⁡2A\frac{1 - \cos 2A}{1 + \cos 2A} = \tan^2A

    L.H.S. =1−(cos⁡2A−sin⁡2A)1+cos⁡2A−sin⁡2A= \frac{1 - (\cos^2A - \sin^2A)}{1 + \cos^2A - \sin^2A}

    =2sin⁡2A2cos⁡2A=tan⁡2A== \frac{2\sin^2A}{2\cos^2A} = \tan^2A = R.H.S.

  7. We have to prove that tan⁡A+cot⁡A=2cosec⁡2A\tan A + \cot A = 2\cosec 2A

    L.H.S. =sin⁡Acos⁡A+cos⁡Asin⁡A=sin⁡2A+cos⁡2Asin⁡Acos⁡A= \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2A + \cos^2A}{\sin A\cos A}

    =22sin⁡Acos⁡A=2sin⁡2A=2cosec⁡2A== \frac{2}{2\sin A\cos A} = \frac{2}{\sin 2A} = 2\cosec 2A = R.H.S.

  8. We have to prove that tan⁡A−cot⁡A=−2cot⁡2A\tan A - \cot A = -2\cot2A

    L.H.S. =sin⁡Acos⁡A−cos⁡Asin⁡A=sin⁡2A−cos⁡2Asin⁡Acos⁡A= \frac{\sin A}{\cos A} - \frac{\cos A}{\sin A} = \frac{\sin^2A - \cos^2A}{\sin A\cos A}

    =−cos⁡2Asin⁡2A2=−2cot⁡2A== \frac{-\cos2A}{\frac{\sin2A}{2}} = -2\cot2A = R.H.S.

  9. We have to prove that cosec⁡2A+cot⁡2A=cot⁡A\cosec 2A + \cot 2A = \cot A

    L.H.S. =1sin⁡2A+cos⁡2Asin⁡2A=1+cos⁡2Asin⁡2A=2cos⁡2A2sin⁡Acos⁡A= \frac{1}{\sin 2A} + \frac{\cos 2A}{\sin 2A} = \frac{1 + \cos 2A}{\sin 2A} = \frac{2\cos^2A}{2\sin A\cos A}

    =cot⁡A== \cot A = R.H.S.

  10. We have to prove that 1−cos⁡A+cos⁡B−cos⁡(A+B)1+cos⁡A−cos⁡B−cos⁡(A+B)=tan⁡A2cot⁡B2\frac{1 - \cos A + \cos B - \cos(A + B)}{1 + \cos A - \cos B - \cos(A + B)} = \tan\frac{A}{2}\cot\frac{B}{2}

    L.H.S. =1−cos⁡A+cos⁡B−cos⁡(A+B)1+cos⁡A−cos⁡B−cos⁡(A+B)= \frac{1 - \cos A + \cos B - \cos(A + B)}{1 + \cos A - \cos B - \cos(A + B)}

    =2sin⁡2A2+2sin⁡A2sin⁡(A2+B)2cos⁡2A2−2cos⁡A2cos⁡(A2+B)= \frac{2\sin^2\frac{A}{2} + 2\sin\frac{A}{2}\sin\left(\frac{A}{2} + B\right)}{2\cos^2\frac{A}{2} - 2\cos\frac{A}{2}\cos\left(\frac{A}{2} + B\right)}

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

    =tan⁡A2(2sin⁡(A+B2)cos⁡B2)2sin⁡(A+B2)sin⁡B2= \frac{\tan\frac{A}{2}\left(2\sin\left(\frac{A + B}{2}\right)\cos\frac{B}{2}\right)}{2\sin\left(\frac{A + B}{2}\right)\sin\frac{B}{2}}

    =tan⁡A2cot⁡B2= \tan\frac{A}{2}\cot\frac{B}{2}

  11. We have to prove that cos⁡A1∓sin⁡A=tan⁡(45∘±A2)\frac{\cos A}{1 \mp \sin A} = \tan\left(45^\circ \pm \frac{A}{2}\right)

    First considering −- sign on L.H.S.,

    L.H.S. =cos⁡A1−sin⁡A=cos⁡2A2−sin⁡2A2(cos⁡A2−sin⁡A2)2= \frac{\cos A}{1 - \sin A} = \frac{\cos^2\frac{A}{2} - \sin^2\frac{A}{2}}{\left(\cos\frac{A}{2} - \sin\frac{A}{2}\right)^2}

    Dividing numerator and denomiator by cos⁡2A2\cos^2\frac{A}{2}

    =1−tan⁡2A2(1−tan⁡A2)2= \frac{1 - \tan^2\frac{A}{2}}{\left(1 - \tan\frac{A}{2}\right)^2}

    =1+tan⁡A21−tan⁡A2= \frac{1 + \tan\frac{A}{2}}{1 - \tan\frac{A}{2}}

    =tan⁡45∘+tan⁡A21−tan⁡45∘tan⁡A2=tan⁡(45∘+A2)= \frac{\tan45^\circ + \tan\frac{A}{2}}{1 - \tan45^\circ\tan\frac{A}{2}} = \tan\left(45^\circ + \frac{A}{2}\right)

    Similarly by considering the ++ sign we can prove the other sign.

  12. We have to prove that sec⁡8A−1sec⁡4A−1=tan⁡8Atan⁡2A\frac{\sec 8A - 1}{\sec 4A - 1} = \frac{\tan 8A}{\tan 2A}

    L.H.S. =sec⁡8A−1sec⁡4A−1=1−cos⁡8A1−cos⁡4A.cos⁡4Acos⁡8A= \frac{\sec 8A - 1}{\sec 4A - 1} = \frac{1 - \cos 8A}{1 - \cos 4A}.\frac{\cos4A}{\cos8A}

    =2sin⁡24A2sin⁡22A.cos⁡4Acos⁡8A=(2sin⁡4Acos⁡4A).sin⁡4A2sin⁡22A.cos⁡8A= \frac{2\sin^24A}{2\sin^22A}.\frac{\cos 4A}{\cos8A} = \frac{(2\sin4A\cos4A).\sin4A}{2\sin^22A.\cos8A}

    =sin⁡8Acos⁡8A.sin⁡4A2sin⁡22A=tan⁡8A.2sin⁡2Acos⁡2A2sin⁡22A=tan⁡8Atan⁡2A== \frac{\sin8A}{\cos8A}.\frac{\sin4A}{2\sin^22A} = \frac{\tan8A. 2\sin2A\cos2A}{2\sin^22A} = \frac{\tan8A}{\tan2A} = R.H.S.

  13. We have to prove that 1+tan⁡2(45∘−A)1−tan⁡2(45∘−A)=cosec⁡2A\frac{1 + \tan^2(45^\circ - A)}{1 - \tan^2(45^\circ - A)} = \cosec 2A

    L.H.S. =1+tan⁡2(45∘−A)1−tan⁡2(45∘−A)= \frac{1 + \tan^2(45^\circ - A)}{1 - \tan^2(45^\circ - A)}

    =cos⁡2(45∘−A)+sin⁡2(45∘−A)cos⁡2(45∘−A)−sin⁡2(45∘−A)= \frac{\cos^2(45^\circ - A) + \sin^2(45^\circ - A)}{\cos^2(45^\circ - A) - \sin^2(45^\circ - A)}

    =1cos⁡(90∘−2A)=1sin⁡2A=cosec⁡2A== \frac{1}{\cos(90^\circ - 2A)} = \frac{1}{\sin2A} = \cosec2A = R.H.S.

  14. We have to prove that sin⁡A+sin⁡Bsin⁡A−sin⁡B=tan⁡A+B2tan⁡A−B2\frac{\sin A + \sin B}{\sin A - \sin B} = \frac{\tan \frac{A + B}{2}}{\tan \frac{A - B}{2}}

    L.H.S. =sin⁡A+sin⁡Bsin⁡A−sin⁡B=2sin⁡A+B2cos⁡A−B22cos⁡A+B2sin⁡A−B2= \frac{\sin A + \sin B}{\sin A - \sin B} = \frac{2\sin\frac{A + B}{2}\cos\frac{A - B}{2}}{2\cos\frac{A + B}{2}\sin\frac{A - B}{2}}

    =tan⁡A+B2tan⁡A−B2== \frac{\tan \frac{A + B}{2}}{\tan \frac{A - B}{2}} = R.H.S.

  15. We have to prove that sin⁡2A−sin⁡2Bsin⁡Acos⁡A−sin⁡Bcos⁡B=tan⁡(A+B)\frac{\sin^2A - \sin^2B}{\sin A\cos A - \sin B\cos B} = \tan(A + B)

    L.H.S. =2(cos⁡2B−cos⁡2A)sin⁡2A−sin⁡2B=cos⁡2B−cos⁡2Asin⁡2A−sin⁡2B= \frac{2(\cos^2B - \cos^2A)}{\sin2A - \sin2B} = \frac{\cos2B - \cos2A}{\sin2A - \sin2B}

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

  16. We have to prove that tan⁡(π4+A)−tan⁡(π4−A)=2tan⁡2A\tan\left(\frac{\pi}{4} + A\right) - \tan\left(\frac{\pi}{4} - A\right) = 2\tan 2A

    L.H.S. =1+tan⁡A1−tan⁡A−1−tan⁡A1+tan⁡A= \frac{1 + \tan A}{1 - \tan A} - \frac{1 - \tan A}{1 + \tan A}

    =(1+tan⁡A)2−(1−tan⁡A)21−tan⁡2A=4tan⁡A1−tan⁡2A= \frac{(1 + \tan A)^2 - (1 - \tan A)^2}{1 - \tan^2A} = \frac{4\tan A}{1 - \tan^2A}

    =4sin⁡Acos⁡A.cos⁡2Acos⁡2A−sin⁡2A=2sin⁡2Acos⁡2A=2tan⁡2A== \frac{4\sin A}{\cos A}. \frac{\cos^2A}{\cos^2A - \sin^2A} = \frac{2\sin2A}{\cos2A} = 2\tan2A = R.H.S.

  17. We have to prove that cos⁡A+sin⁡Acos⁡A−sin⁡A−cos⁡A−sin⁡Acos⁡A+sin⁡A=2tan⁡2A\frac{\cos A + \sin A}{\cos A - \sin A} - \frac{\cos A - \sin A}{\cos A + \sin A} = 2\tan 2A

    L.H.S. =(cos⁡A+sin⁡A)2−(cos⁡A−sin⁡A)2cos⁡2A−sin⁡2A= \frac{(\cos A + \sin A)^2 - (\cos A - \sin A)^2}{\cos^2A - \sin^2A}

    =4cos⁡Asin⁡Acos⁡2A=2sin⁡2Acos⁡2A=2tan⁡2A== \frac{4\cos A\sin A}{\cos 2A} = \frac{2\sin 2A}{\cos 2A} = 2\tan2A = R.H.S.

  18. We have to prove that cot⁡(A+15∘)−tan⁡(A−15∘)=4cos⁡2A1+2sin⁡2A\cot (A + 15^\circ) - \tan(A - 15^\circ) = \frac{4\cos 2A}{1 + 2\sin 2A}

    L.H.S. =1−tan⁡(A+15∘)tan⁡(A−15∘)tan⁡(A+15∘)= \frac{1 - \tan(A + 15^\circ)\tan(A - 15^\circ)}{\tan(A + 15^\circ)}

    =cos⁡(A+15∘)cos⁡(A−15∘)−sin⁡(A+15∘)sin⁡(A−15∘)cos⁡(A+15∘)cos⁡(A−15∘).cos⁡(A+15∘)sin⁡(A+15∘)= \frac{\cos(A + 15^\circ)\cos(A - 15^\circ) - \sin(A + 15^\circ)\sin(A - 15^\circ)}{\cos(A + 15^\circ)\cos(A - 15^\circ)}.\frac{\cos(A + 15^\circ)}{\sin(A + 15^\circ)}

    =cos⁡2Asin⁡(A+15∘)cos⁡(A−15∘)=2cos⁡2A2sin⁡(A+15∘)cos⁡(A−15∘)= \frac{\cos 2A}{\sin(A + 15^\circ)\cos(A - 15^\circ)} = \frac{2\cos 2A}{2\sin(A + 15^\circ)\cos(A - 15^\circ)}

    =2cos⁡2Asin⁡2A+sin⁡30∘=4cos⁡2A1+sin⁡2A== \frac{2\cos 2A}{\sin2A + \sin30^\circ} = \frac{4\cos 2A}{1 + \sin 2A} = R.H.S.

  19. We have to prove that sin⁡A+sin⁡2A1+cos⁡A+cos⁡2A=tan⁡A\frac{\sin A + \sin2A}{1 + \cos A + \cos 2A} = \tan A

    L.H.S. =sin⁡A+2sin⁡Acos⁡Acos⁡A+2cos⁡2A=sin⁡A(1+2cos⁡A)cos⁡A(1+2cos⁡A)= \frac{\sin A + 2\sin A\cos A}{\cos A + 2\cos^2A} = \frac{\sin A(1 + 2\cos A)}{\cos A(1 + 2\cos A)}

    =tan⁡A== \tan A = R.H.S.

  20. We have to prove that 1+sin⁡A−cos⁡A1+sin⁡A+cosA=tan⁡A2\frac{1 + \sin A - \cos A }{1 + \sin A + cos A} = \tan \frac{A}{2}

    L.H.S. =2sin⁡2A2+2sin⁡A2cos⁡A22cos⁡2A2+2sin⁡A2cos⁡A2= \frac{2\sin^2\frac{A}{2} + 2\sin\frac{A}{2}\cos\frac{A}{2}}{2\cos^2\frac{A}{2} + 2\sin\frac{A}{2}\cos\frac{A}{2}}

    =sin⁡A2(sin⁡A2+cos⁡A2)cos⁡A2(sin⁡A2+cos⁡A2)= \frac{\sin\frac{A}{2}(\sin\frac{A}{2} + \cos\frac{A}{2})}{\cos\frac{A}{2}(\sin\frac{A}{2} + \cos\frac{A}{2})}

    =tan⁡A2== \tan\frac{A}{2} = R.H.S.

  21. We have to prove that sin⁡(n+1)A−sin⁡(n−1)Acos⁡(n+1)A+2cos⁡nA+cos⁡(n−1)A=tan⁡A2\frac{\sin(n + 1)A - \sin(n - 1)A}{\cos(n + 1)A + 2\cos nA + \cos(n - 1)A} = \tan \frac{A}{2}

    L.H.S. =2cos⁡nAsin⁡A2cos⁡nAcos⁡A+2cos⁡nA=sin⁡A1+cos⁡A= \frac{2\cos nA \sin A}{2\cos nA \cos A + 2\cos nA} = \frac{\sin A}{1 + \cos A}

    =2sin⁡A2cos⁡A22cos⁡2A2=tan⁡A2== \frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{2\cos^2\frac{A}{2}} = \tan \frac{A}{2} = R.H.S.

  22. We have to prove that sin⁡(n+1)A+2sin⁡nA+sin⁡(n−1)Acos⁡(n−1)−cos⁡(n+1)A=cot⁡A2\frac{\sin(n + 1)A + 2\sin nA + \sin(n - 1)A}{\cos(n - 1) - \cos(n + 1)A} = \cot \frac{A}{2}

    L.H.S. =2sin⁡nAcos⁡A+2sin⁡nA2sin⁡nAsin⁡A= \frac{2\sin nA\cos A + 2\sin nA}{2\sin nA\sin A}

    =cos⁡A+1sin⁡A=2cos⁡2A22sin⁡A2cos⁡A2= \frac{\cos A + 1}{\sin A} = \frac{2\cos^2\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}}

    =cot⁡A2== \cot\frac{A}{2} = R.H.S.

  23. We have to prove that sin⁡(2n+1)Asin⁡A=sin⁡2(n+1)A−sin⁡2nA\sin(2n + 1)A\sin A = \sin^2(n + 1)A - \sin^2nA

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

    =(2sin⁡2n+12Acos⁡A2)(2cos⁡2n+12Asin⁡A2)= (2\sin\frac{2n + 1}{2}A\cos \frac{A}{2})(2\cos\frac{2n + 1}{2}A\sin \frac{A}{2})

    =2sin⁡2n+12Acos⁡2n+12A.2cos⁡A2sin⁡A2= 2\sin\frac{2n + 1}{2}A\cos\frac{2n + 1}{2}A.2\cos \frac{A}{2}\sin\frac{A}{2}

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

  24. We have to prove that sin⁡(A+3B)+sin⁡(3A+B)sin⁡2A+sin⁡2B=2cos⁡(A+B)\frac{\sin(A + 3B) + \sin(3A + B)}{\sin 2A + \sin 2B} = 2\cos(A + B)

    L.H.S. =sin⁡(A+3B)+sin⁡(3A+B)sin⁡2A+sin⁡2B= \frac{\sin(A + 3B) + \sin(3A + B)}{\sin 2A + \sin 2B}

    =2sin⁡(2A+2B)cos⁡(A−B)2sin⁡(A+B)cos⁡(A−B)= \frac{2\sin(2A + 2B)\cos(A - B)}{2\sin(A + B)\cos(A - B)}

    =2sin⁡(A+B)cos⁡(A+B)sin⁡(A+B)=2cos⁡(A+B)== \frac{2\sin(A + B)\cos(A + B)}{\sin(A + B)} = 2\cos(A + B) = R.H.S.

  25. We have to prove that sin⁡3A+sin⁡2A−sin⁡A=4sin⁡Acos⁡A2cos⁡3A2\sin 3A + \sin 2A - \sin A = 4\sin A\cos \frac{A}{2}\cos \frac{3A}{2}

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

    =2sin⁡Acos⁡3A2cos⁡A2== 2\sin A\cos \frac{3A}{2}\cos\frac{A}{2} = R.H.S.

  26. We have to prove that tan⁡2A=(sec⁡2A+1)sec⁡2A−1\tan 2A = (\sec 2A + 1)\sqrt{\sec^2A - 1}

    R.H.S. =1+cos⁡2Acos⁡2A1−cos⁡2Acos⁡2A= \frac{1 + \cos 2A}{\cos 2A}\sqrt{\frac{1 - \cos^2A}{\cos^2A}}

    =2cos⁡2A2cos⁡2A−1.sin⁡2Acos⁡2A= \frac{2\cos^2A}{2\cos^2A - 1}.\sqrt{\frac{\sin^2A}{\cos^2A}}

    =22−sec⁡2A.tanA=2tan⁡A1−tan⁡2A=tan⁡A+tan⁡A1−tan⁡A.tan⁡A= \frac{2}{2 - \sec^2A}.tan A = \frac{2\tan A}{1 - \tan^2A} = \frac{\tan A + \tan A}{1 - \tan A.\tan A}

    =tan⁡2A==\tan 2A = R.H.S.

  27. We have to prove that cos⁡32A+3cos⁡2A=4(cos⁡6A−sin⁡6A)\cos^32A + 3\cos 2A = 4(\cos^6A - \sin^6A)

    L.H.S. =(cos⁡2A−sin⁡2A)3+3(cos⁡2A−sin⁡2A)= (\cos^2A - \sin^2A)^3 + 3(\cos^2A - \sin^2A)

    =cos⁡6A−3cos⁡4Asin⁡2A+3cos⁡2Asin⁡4A−sin⁡6A+3(cos⁡2A−sin⁡2A)= \cos^6A -3\cos^4A\sin^2A + 3\cos^2A\sin^4A - \sin^6A + 3(\cos^2A - \sin^2A)

    =cos⁡6A−3cos⁡4A(1−cos⁡2A)+3(1−sin⁡2A)sin⁡4A−sin⁡6A+3(cos⁡2A−sin⁡2A)= \cos^6A -3\cos^4A(1 - \cos^2A) + 3(1 - \sin^2A)\sin^4A - \sin^6A + 3(\cos^2A - \sin^2A)

    =4(cos⁡6A−sin⁡6A)== 4(\cos^6A - \sin^6A) = R.H.S.

  28. We have to prove that 1+cos⁡22A=2(cos⁡4A+sin⁡4A)1 + \cos^22A = 2(\cos^4A + \sin^4A)

    L.H.S. =1+(cos⁡2A−sin⁡2A)2=1−2sin⁡2Acos⁡2A+cos⁡4A+sin⁡4A= 1 + (\cos^2A - \sin^2A)^2 = 1 - 2\sin^2A\cos^2A + \cos^4A + \sin^4A

    =1−2sin⁡2A(1−sin⁡2A)+cos⁡4A+sin⁡4A= 1 - 2\sin^2A(1 - \sin^2A) + \cos^4A + \sin^4A

    =1−2sin⁡2A+2sin⁡4A+cos⁡4A+sin⁡4A= 1 - 2\sin^2A + 2\sin^4A + \cos^4A + \sin^4A

    =(1−sin⁡2A)2+cos⁡4A+2sin⁡4A=2(cos⁡4A+sin⁡4A)== (1 - \sin^2A)^2 + \cos^4A + 2\sin^4A = 2(\cos^4A + \sin^4A) = R.H.S.

  29. We have to prove that sec⁡2A(1+sec⁡2A)=2sec⁡2A\sec^2A(1 + \sec2A) = 2\sec2A

    L.H.S. =1cos⁡2A.cos⁡2A+1cos⁡2A= \frac{1}{\cos^2A}.\frac{\cos2A + 1}{\cos 2A}

    =1cos⁡2A.2cos⁡2Acos⁡2A=2sec⁡2A== \frac{1}{\cos^2A}.\frac{2\cos^2A}{\cos 2A} = 2\sec2A = R.H.S.

  30. We have to prove that cosec⁡A−2cot⁡2Acos⁡A=2sin⁡A\cosec A - 2\cot 2A\cos A = 2\sin A

    L.H.S. =1sin⁡A−2cos⁡2Acos⁡Asin⁡2A= \frac{1}{\sin A} - \frac{2\cos 2A\cos A}{\sin 2A}

    =1sin⁡A−2cos⁡2Acos⁡A2sin⁡Acos⁡A= \frac{1}{\sin A} - \frac{2\cos 2A\cos A}{2\sin A\cos A}

    1sin⁡A−cos⁡2Asin⁡A=1−cos⁡2Asin⁡A\frac{1}{\sin A} - \frac{\cos 2A}{\sin A} = \frac{1 - \cos 2A}{\sin A}

    =2sin⁡2Asin⁡A=2sin⁡A== \frac{2\sin^2A}{\sin A} = 2\sin A = R.H.S.

  31. We have to prove that cot⁡A=12(cot⁡A2−tan⁡A2)\cot A = \frac{1}{2}\left(\cot\frac{A}{2} - \tan\frac{A}{2}\right)

    R.H.S. =12(1−tan⁡2A2tan⁡A2)= \frac{1}{2}\left(\frac{1 - \tan^2\frac{A}{2}}{\tan\frac{A}{2}}\right)

    =12(cos⁡2A2−sin⁡2A2cos⁡2A2).cos⁡A2sin⁡A2= \frac{1}{2}\left(\frac{\cos^2\frac{A}{2} - \sin^2\frac{A}{2}}{\cos^2\frac{A}{2}}\right).\frac{\cos\frac{A}{2}}{\sin \frac{A}{2}}

    =12cos⁡Acos⁡A2.1sin⁡A2=cot⁡A== \frac{1}{2}\frac{\cos A}{\cos\frac{A}{2}}.\frac{1}{\sin\frac{A}{2}} = \cot A = L.H.S.

  32. We have to prove that sin⁡Asin⁡(60∘−A)sin⁡(60∘+A)=14sin⁡3A\sin A\sin(60^\circ - A)\sin(60^\circ + A) = \frac{1}{4}\sin 3A

    L.H.S. =sin⁡A.cos⁡2A−cos⁡120∘2=sin⁡A(1−2sin⁡2A+12)2=\sin A.\frac{\cos 2A - \cos 120^\circ}{2} = \frac{\sin A\left(1 - 2\sin^2A + \frac{1}{2}\right)}{2}

    =3sin⁡A−4sin⁡3A4=14sin⁡3A== \frac{3\sin A - 4\sin^3A}{4} = \frac{1}{4}\sin 3A = R.H.S.

  33. We have to prove that cos⁡Acos⁡(60∘−A)cos⁡(60∘+A)=14cos⁡3A\cos A\cos(60^\circ - A)\cos(60^\circ + A) = \frac{1}{4}\cos 3A

    L.H.S. =cos⁡A2(cos⁡2A+cos⁡120∘)=cos⁡A2(2cos⁡2A−1−12)= \frac{\cos A}{2}\left(\cos 2A + \cos120^\circ\right) = \frac{\cos A}{2}\left(2\cos^2A - 1 - \frac{1}{2}\right)

    =4cos⁡3A−3cos⁡A4=14cos⁡3A== \frac{4\cos^3A - 3\cos A}{4} = \frac{1}{4}\cos 3A = R.H.S.

  34. We have to prove that cot⁡A+cot⁡(60∘+A)−cot⁡(60∘−A)=3cot⁡3A\cot A + \cot(60^\circ + A) - \cot(60^\circ - A) = 3\cot 3A

    L.H.S. =1tan⁡A+1tan⁡(60∘+A)−1tan⁡(60∘−A)= \frac{1}{\tan A} + \frac{1}{\tan(60^\circ + A)} - \frac{1}{\tan(60^\circ - A)}

    =1tan⁡A+1−3tan⁡A3+tan⁡A−1+3tan⁡A3−tan⁡A= \frac{1}{\tan A} + \frac{1 - \sqrt{3}\tan A}{\sqrt{3} + \tan A} - \frac{1 + \sqrt{3}\tan A}{\sqrt{3} - \tan A}

    =1tan⁡A−8tan⁡A3−tan⁡2A=3(1−3tan⁡2A)3tan⁡A−tan⁡3A=3tan⁡3A= \frac{1}{\tan A} - \frac{8\tan A}{3 - \tan^2A} = \frac{3(1 - 3\tan^2A)}{3\tan A - \tan^3A} = \frac{3}{\tan 3A}

    =3cot⁡3A== 3\cot 3A = R.H.S.

  35. We have to prove that cos⁡4A=1−8cos⁡2A+8cos⁡4A\cos 4A = 1 - 8\cos^2A + 8\cos^4A

    L.H.S. =cos⁡4A=2cos⁡22A−1=2(2cos⁡2A−1)2−1= \cos 4A = 2\cos^22A - 1 = 2(2\cos^2A - 1)^2 - 1

    =2(4cos⁡4A−4cos⁡2A+1)−1=2(4\cos^4A - 4\cos^2A + 1) - 1

    =1−8cos⁡2A+8cos⁡4A== 1 - 8\cos^2A + 8\cos^4A = R.H.S.

  36. We have to prove that sin⁡4A=4sin⁡Acos⁡3A−4cos⁡Asin⁡3A\sin 4A = 4\sin A\cos^3A - 4\cos A\sin^3A

    L.H.S. =2sin⁡2Acos⁡2A=4sin⁡Acos⁡A(cos⁡2A−sin⁡2A)= 2\sin 2A\cos 2A = 4\sin A\cos A(\cos^2A - \sin^2A)

    =4sin⁡Acos⁡3A−4cos⁡Asin⁡3A== 4\sin A\cos^3A - 4\cos A\sin^3A = R.H.S.

  37. We have to prove that cos⁡6A=32cos⁡6A−48cos⁡4A+18cos⁡2A−1\cos 6A = 32\cos^6A - 48\cos^4A + 18\cos^2A - 1

    L.H.S. =cos⁡6A=(cos⁡23A−sin⁡23A)=(4cos⁡3A−3cos⁡A)2−(3sin⁡A−4sin⁡3A)2= \cos 6A = (\cos^23A - \sin^23A) = (4\cos^3A - 3\cos A)^2 - (3\sin A - 4\sin^3A)^2

    =16cos⁡6A+9cos⁡2A−24cos⁡4A−9sin⁡2A−16sin⁡6A+24sin⁡4A= 16\cos^6A + 9\cos^2A -24\cos^4A - 9\sin^2A - 16\sin^6A + 24\sin^4A

    =16cos⁡6A+9cos⁡2A−24cos⁡4A−9(1−cos⁡2A)−16(1−cos⁡2A)3+24(1−cos⁡2A)2= 16\cos^6A + 9\cos^2A -24\cos^4A - 9(1 - \cos^2A) - 16(1 - \cos^2A)^3 + 24(1 - \cos^2A)^2

    =32cos⁡6A−48cos⁡4A+18cos⁡2A−1== 32\cos^6A - 48\cos^4A + 18\cos^2A - 1 = R.H.S.

  38. We have to prove that tan⁡3Atan⁡2Atan⁡A=tan⁡3A−tan⁡2A−tan⁡A\tan 3A\tan 2A\tan A = \tan 3A - \tan 2A - \tan A

    Rewriting this as following:

    tan⁡A+tan⁡2A=tan⁡3A(1−tan⁡Atan⁡2A)⇒tan⁡A+tan⁡2A1−tan⁡Atan⁡2A=tan⁡3A\tan A + \tan 2A = \tan 3A(1 - \tan A\tan 2A)\Rightarrow \frac{\tan A + \tan 2A}{1 - \tan A\tan 2A} = \tan 3A

    ⇒tan⁡(A+2A)=tan⁡3A\Rightarrow \tan (A + 2A) = \tan 3A

    Hence, proved.

  39. We have to prove that 2cos⁡2nA+12cos⁡A+1=(2cos⁡A−1)(2cos⁡2A−1)(2cos⁡22A−1)…(2cos⁡2n−1−1)\frac{2\cos2^nA + 1}{2\cos A + 1} = (2\cos A - 1)(2\cos 2A - 1)(2\cos2^2A - 1)\ldots(2\cos2^{n - 1} - 1)

    L.H.S. =2cos⁡2nA+12cos⁡A+1= \frac{2\cos2^nA + 1}{2\cos A + 1}

    Multiplying and dividing by 2cos⁡A−12\cos A - 1

    =(2cos⁡A−1)2cos⁡2nA+14cos⁡2A−1=(2cos⁡A−1)2cos⁡2nA+12cos⁡2A+1= (2\cos A - 1)\frac{2\cos2^nA + 1}{4\cos^2A - 1} = (2\cos A - 1)\frac{2\cos2^nA + 1}{2\cos2A + 1}

    Multiplying and dividing by 2cos⁡2A−12\cos2A - 1

    =(2cos⁡A−1)(2cos⁡2A−1)2cos⁡2nA+14cos⁡22A−1= (2\cos A - 1)(2\cos2A - 1)\frac{2\cos2^nA + 1}{4\cos^22A - 1}

    =(2cos⁡A−1)(2cos⁡2A−1)2cos⁡2nA+12cos⁡22A+1= (2\cos A - 1)(2\cos2A - 1)\frac{2\cos2^nA + 1}{2\cos2^2A + 1}

    Proceeding similarly we obtain the R.H.S.

  40. Given tan⁡A=17,sin⁡B=110\tan A= \frac{1}{7}, \sin B = \frac{1}{\sqrt{10}}

    ∴cos⁡B=310,tan⁡B=13\therefore \cos B = \frac{3}{\sqrt{10}}, \tan B = \frac{1}{3}

    tan⁡(A+2B)=tan⁡A+tan⁡2B1−tan⁡Atan⁡2B\tan(A + 2B) = \frac{\tan A + \tan 2B}{1 - \tan A\tan 2B}

    =tan⁡A+2tan⁡B1−tan⁡2B1−tan⁡A.2tan⁡B1−tan⁡2B= \frac{\tan A + \frac{2\tan B}{1 - \tan^2B}}{1 - \tan A.\frac{2\tan B}{1 - \tan^2B}}

    =17+2131−191−17.2131−19= \frac{\frac{1}{7} + \frac{2\frac{1}{3}}{1 - \frac{1}{9}}}{1 - \frac{1}{7}.\frac{2\frac{1}{3}}{1 - \frac{1}{9}}}

    =1∴A+2B=π4= 1 \therefore A + 2B = \frac{\pi}{4}

  41. We have to prove that tan⁡(π4+A)+tan⁡(π4−A)=2sec⁡2A\tan\left(\frac{\pi}{4} + A\right) + \tan\left(\frac{\pi}{4} - A\right) = 2\sec2A

    L.H.S. =1+tan⁡A1−tan⁡A+1−tan⁡A1+tan⁡A= \frac{1 + \tan A}{1 - \tan A} + \frac{1 - \tan A}{1 + \tan A}

    =(1+tan⁡A)2+(1−tan⁡A)21−tan⁡2A=2+2tan⁡2A1−tan⁡2A= \frac{(1 + \tan A)^2 + (1 - \tan A)^2}{1 - \tan^2A} = \frac{2 + 2\tan^2A}{1 - \tan^2A}

    =2(sin⁡2A+cos⁡2A)cos⁡2A−sin⁡2A=2cos⁡2A=2sec⁡2A== \frac{2(\sin^2A + \cos^2A)}{\cos^2A - \sin^2A} = \frac{2}{\cos 2A} = 2\sec 2A = R.H.S.

  42. We have to prove that 3cosec⁡20∘−sec⁡20∘=4\sqrt{3}\cosec 20^\circ - \sec 20^\circ = 4

    L.H.S. =3sin⁡20∘−1cos⁡20∘= \frac{\sqrt{3}}{\sin20^\circ} - \frac{1}{\cos20^\circ}

    =4(32)cos⁡20∘−12sin⁡20∘2sin⁡20∘cos⁡20∘= \frac{4(\frac{\sqrt{3}}{2})\cos20^\circ - \frac{1}{2}\sin20^\circ}{2\sin20^\circ\cos^20\circ}

    =4(sin⁡(50∘−20∘))sin⁡40∘=4== \frac{4(\sin(50^\circ - 20^\circ))}{\sin40^\circ} = 4 = R.H.S.

  43. We have to prove that tan⁡A+2tan⁡2A+4tan⁡4A+8cot⁡8A=cot⁡A\tan A + 2\tan 2A + 4\tan 4A + 8\cot 8A = \cot A

    tan⁡A−cot⁡A=sin⁡2A−cos⁡2Asin⁡Acos⁡A=−2cos⁡2Asin⁡2A=−2cot⁡2A\tan A - \cot A = \frac{\sin^2A - \cos^2A}{\sin A\cos A} = -\frac{2\cos 2A}{\sin 2A} = -2\cot 2A

    Similarly, 2tan⁡2A−2cot⁡2A=−4cot⁡4A2\tan 2A - 2\cot 2A = -4\cot 4A

    and 4tan⁡4A−4cot⁡4A=−8cot⁡8A4\tan 4A - 4\cot 4A = -8\cot 8A

    Thus, tan⁡A+2tan⁡2A+4tan⁡4A+8cot⁡8A=cot⁡A\tan A + 2\tan 2A + 4\tan 4A + 8\cot 8A = \cot A

  44. We have to prove that cos⁡2A+cos⁡2(2π3−A)+cos⁡2(2π3+A)=32\cos^2A + \cos^2\left(\frac{2\pi}{3} - A\right) + \cos^2\left(\frac{2\pi}{3} + A\right) = \frac{3}{2}

    ⇒2cos⁡2A+2cos⁡2(2π3−A)+2cos⁡2(2π3+A)=3\Rightarrow 2\cos^2A + 2\cos^2\left(\frac{2\pi}{3} - A\right) + 2\cos^2\left(\frac{2\pi}{3} + A\right) = 3

    L.H.S. =cos⁡2A+1+cos⁡(4π3−2A)+1+cos⁡(4π3+2A)+1= \cos 2A + 1 + \cos\left(\frac{4\pi}{3} - 2A\right) + 1 + \cos\left(\frac{4\pi}{3} + 2A\right) + 1

    =3+cos⁡2A+2cos⁡(4π3)cos⁡2A=3== 3 + \cos2A + 2\cos\left(\frac{4\pi}{3}\right)\cos2A = 3 = R.H.S.

  45. 2sin⁡2A+4cos⁡(A+B)sin⁡Asin⁡B+cos⁡2(A+B)2\sin^2A + 4\cos (A + B)\sin A\sin B + \cos2(A + B)

    =2sin⁡2A+2cos⁡(A+B)2sin⁡Asin⁡B+cos⁡2(A+B)= 2\sin^2A + 2\cos(A + B)2\sin A\sin B + \cos2(A + B)

    =2sin⁡2A+2cos⁡(A+B)[cos⁡(A−B)−cos⁡(A+B)]+cos⁡2(A+B)= 2\sin^2A + 2\cos(A + B)[\cos(A - B) - \cos(A + B)] + \cos2(A + B)

    =2sin⁡2A+2cos⁡(A+B)cos⁡(A−B)−2cos⁡2(A+B)+cos⁡2(A+B)= 2\sin^2A + 2\cos(A + B)\cos(A - B) - 2\cos^2(A + B) + \cos2(A + B)

    =2sin⁡2A+2(cos⁡2A−sin⁡2B)−2cos⁡2(A+B)+2cos⁡2(A+B)−1= 2\sin^2A + 2(\cos^2A - \sin^2B) - 2\cos^2(A + B) + 2\cos^2(A + B) - 1

    =2(sin⁡2A+cos⁡2A)−2sin⁡2B−1=1−2sin⁡2B= 2(\sin^2A + \cos^2A) -2\sin^2B - 1 = 1 -2\sin^2B which is independent of AA

  46. Given, cos⁡A=12(a+1a)\cos A = \frac{1}{2}\left(a + \frac{1}{a}\right)

    cos⁡2A=2cos⁡2A−1=2.14(a+1a)2−1\cos 2A = 2\cos^2A - 1 = 2.\frac{1}{4}\left(a + \frac{1}{a}\right)^2 - 1

    =12(a2+1a2)= \frac{1}{2}\left(a^2 + \frac{1}{a^2}\right)

  47. We have to prove that cos⁡2A+sin⁡2Acos⁡2B=cos⁡2B+sin⁡2Bcos⁡2A\cos^2A + \sin^2A\cos 2B = \cos^2B + \sin^2B\cos 2A

    ⇒cos⁡2A−cos⁡2B=sin⁡2Bcos⁡2A−sin⁡2Acos⁡2B\Rightarrow \cos^2A - \cos^2B = \sin^2B\cos2A - \sin^2A\cos2B

    R.H.S. =sin⁡2Bcos⁡2A−sin⁡2Acos⁡2B= \sin^2B\cos2A - \sin^2A\cos2B

    =sin⁡2B(cos⁡2A−sin⁡2A)−sin⁡2A(cos⁡2B−sin⁡2B)= \sin^2B(\cos^2A - \sin^2A) - \sin^2A(\cos^2B - \sin^2B)

    =cos⁡2Asin⁡2B−sin⁡2Acos⁡2B=cos⁡2A(1−cos⁡2B)−(1−cos⁡2A)cos⁡2B= \cos^2A\sin^2B - \sin^2A\cos^2B = \cos^2A(1 - \cos^2B) - (1 - \cos^2A)\cos^2B

    =cos⁡2A−cos⁡2B== \cos^2A - \cos^2B = R.H.S.

  48. We have to prove that 1+tan⁡Atan⁡2A=sec⁡2A1 + \tan A\tan 2A = \sec 2A

    L.H.S. =1+tan⁡Atan⁡2A=1+tan⁡A.2tan⁡A1−tan⁡2A= 1 + \tan A\tan 2A = 1 + \tan A.\frac{2\tan A}{1 - \tan^2A}

    =1+tan⁡2A1−tan⁡2A=cos⁡2A+sin⁡2Acos⁡2A−sin⁡2A= \frac{1 + \tan^2A}{1 - \tan^2A} = \frac{\cos^2A + \sin^2A}{\cos^2A - \sin^2A}

    =1cos⁡2A=sec⁡2A== \frac{1}{\cos 2A} = \sec 2A = R.H.S.

  49. We have to prove that 1+sin⁡2A1−sin⁡2A=(1+tan⁡A1−tan⁡A)2\frac{1 + \sin 2A}{1 - \sin 2A} = \left(\frac{1 + \tan A}{1 - \tan A}\right)^2

    L.H.S. =1+sin⁡2A1−sin⁡2A=sin⁡2A+cos⁡2A+2sin⁡Acos⁡Asin⁡2A+cos⁡2A−2sin⁡Acos⁡A= \frac{1 + \sin 2A}{1 - \sin 2A} = \frac{\sin^2A + \cos^2A + 2\sin A\cos A}{\sin^2A + \cos^2A - 2\sin A\cos A}

    =(sin⁡A+cos⁡Asin⁡A−cos⁡A)2= \left(\frac{\sin A + \cos A}{\sin A - \cos A}\right)^2

    Dividing numerator and denominator by cos⁡2A,\cos^2A, we get

    =(1+tan⁡A1−tan⁡A)2== \left(\frac{1 + \tan A}{1 - \tan A}\right)^2 = R.H.S.

  50. We have to prove that 1sin⁡10∘−3cos⁡10∘=4\frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ} = 4

    L.H.S. =1sin⁡10∘−3cos⁡10∘= \frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ}

    =cos⁡10∘−3sin⁡10∘sin⁡10∘cos⁡10∘= \frac{\cos10^\circ - \sqrt{3}\sin10^\circ}{\sin10^\circ\cos10^\circ}

    =2.2(12cos⁡10∘−32sin⁡10∘)2sin⁡10∘cos⁡10∘= \frac{2.2\left(\frac{1}{2}\cos10^\circ - \frac{\sqrt{3}}{2}\sin10^\circ\right)}{2\sin10^\circ\cos10^\circ}

    =4.sin⁡30∘cos⁡10∘−cos⁡30∘sin⁡10∘sin⁡20∘=4.sin⁡(30∘−10∘)sin⁡20∘= 4.\frac{\sin30^\circ\cos10^\circ - \cos30^\circ\sin10^\circ}{\sin20^\circ} = 4.\frac{\sin(30^\circ - 10^\circ)}{\sin20^\circ}

    =4== 4 = R.H.S.

  51. We have to prove that cot⁡2A−tan⁡2A=4cot⁡2Acosec⁡2A\cot^2A - \tan^2A = 4\cot2A\cosec 2A

    L.H.S. =cos⁡2Asin⁡2A−sin⁡2Acos⁡2A= \frac{\cos^2A}{\sin^2A} - \frac{\sin^2A}{\cos^2A}

    =cos⁡4A−sin⁡4Asin⁡2Acos⁡2A=4(cos⁡2A+sin⁡2A)(cos⁡2A−sin⁡2AA)(2sin⁡Acos⁡A)2= \frac{\cos^4A - \sin^4A}{\sin^2A\cos^2A} = \frac{4(\cos^2A + \sin^2A)(\cos^2A - \sin^2AA)}{(2\sin A\cos A)^2}

    =4cos⁡2Asin⁡22A=4cot⁡2Acosec⁡2A== \frac{4\cos 2A}{\sin^22A} = 4\cot 2A\cosec 2A = R.H.S.

  52. We have to prove that 1+sin⁡2Acos⁡2A=cos⁡A+sin⁡Acos⁡A−sin⁡A=tan⁡(π4+A)\frac{1 +\sin 2A}{\cos2A} = \frac{\cos A + \sin A}{\cos A - \sin A} = \tan\left(\frac{\pi}{4} + A\right)

    L.H.S. =1+sin⁡2Acos⁡2A=sin⁡2A+cos⁡2A+2sin⁡Acos⁡Acos⁡2A−sin⁡2A= \frac{1 +\sin 2A}{\cos2A} = \frac{\sin^2A + \cos^2A + 2\sin A\cos A}{\cos^2A - \sin^2A}

    =(cos⁡A+sin⁡A)2cos⁡2A−sin⁡2A=cos⁡A+sin⁡Acos⁡A−sin⁡A== \frac{(\cos A + \sin A)^2}{\cos^2A - \sin^2A} = \frac{\cos A + \sin A}{\cos A - \sin A} = middle term

    Dividing both numerator and denominator by cos⁡A,\cos A, we get

    =1+tan⁡A1−tan⁡A=tan⁡π4+tan⁡A1−tan⁡π4.tan⁡A= \frac{1 + \tan A}{1 - \tan A} = \frac{\tan\frac{\pi}{4} + \tan A}{1 - \tan\frac{\pi}{4}.\tan A}

    =tan⁡(π4+A)== \tan\left(\frac{\pi}{4} + A\right) = R.H.S.

  53. We have to prove that cos⁡6A−sin⁡6A=cos⁡2A(1−14sin⁡22A)\cos^6A - \sin^6A = \cos2A\left(1 - \frac{1}{4}\sin^22A\right)

    R.H.S. =cos⁡2A(1−14sin⁡22A)=(cos⁡2A−sin⁡2A)(1−sin⁡2Acos⁡2A)= \cos2A\left(1 - \frac{1}{4}\sin^22A\right) = (\cos^2A - \sin^2A)(1 - \sin^2A\cos^2A)

    =(cos⁡2A−sin⁡2A)[(cos⁡2A+sin⁡2A)2−sin⁡2Acos⁡2A]=cos⁡6A−sin⁡6A== (\cos^2A - \sin^2A)[(\cos^2A + \sin^2A)^2 - \sin^2A\cos^2A] = \cos^6A - \sin^6A = L.H.S.

  54. This problem is similar to 44 and can be solved similarly.

  55. We have to prove that (1+sec⁡2A)(1+sec⁡22A)(1+sec23A)…(1+sec⁡2nA)=tan⁡2nAtan⁡A(1 + \sec2A)(1+ \sec2^2A)(1 + sec2^3A) \ldots (1 + \sec2^nA) = \frac{\tan2^nA}{\tan A}

    L.H.S. =(1+sec⁡2A)(1+sec⁡22A)(1+sec⁡23A)…(1+sec⁡2nA)= (1 + \sec2A)(1 + \sec2^2A)(1 + \sec2^3A) \ldots (1 + \sec2^nA)

    =tan⁡Atan⁡A(1+sec⁡2A)(1+sec⁡22A)(1+sec⁡23A)…(1+sec⁡2nA)= \frac{\tan A}{\tan A}(1 + \sec2A)(1 + \sec2^2A)(1 + \sec2^3A) \ldots (1 + \sec2^nA)

    Now tan⁡A(1+sec⁡2A)=tan⁡A1+cos⁡2Acos⁡2A\tan A(1 + \sec 2A) = \tan A\frac{1 + \cos 2A}{\cos 2A}

    =tan⁡A1+1−tan⁡2A1+tan⁡2A1−tan⁡2A1+tan⁡2A= \tan A\frac{1 + \frac{1 - \tan^2A}{1 + \tan^2A}}{\frac{1 - \tan^2A}{1 + \tan^2A}}

    =tan⁡A21−tan⁡2A=2tan⁡A1−tan⁡2A=tan⁡2A= \tan A\frac{2}{1 - \tan^2A} = \frac{2\tan A}{1 - \tan^2A} = \tan 2A

    Similarly, tan⁡2A(1+sec⁡22A)=tan⁡22A\tan 2A(1 + \sec2^2A) = \tan2^2A

    Proceeding similalry we obtain R.H.S.

  56. We have to prove that sin⁡2nAsin⁡A=2ncos⁡Acos⁡2Acos⁡22A…cos⁡2n−1A\frac{\sin2^nA}{\sin A} = 2^n\cos A\cos 2A\cos 2^2A\ldots\cos2^{n - 1}A

    Dividing and multiplying with 2cos⁡A2\cos A

    L.H.S. =sin⁡2nAsin⁡A=2cos⁡A.sin⁡2nA2sin⁡Acos⁡A=2cos⁡A.sin⁡2nAsin⁡2A= \frac{\sin2^nA}{\sin A} = 2\cos A.\frac{\sin2^nA}{2\sin A\cos A} = 2\cos A.\frac{\sin2^nA}{\sin 2A}

    Again, dividing and multiplying with 2cos⁡2A2\cos 2A

    =22cos⁡Acos⁡2A.sin⁡2nA2sin⁡2Acos⁡2A=22cos⁡Acos⁡2A.sin⁡2nAsin⁡22A= 2^2\cos A\cos 2A.\frac{\sin2^n A}{2\sin 2A\cos 2A} = 2^2\cos A\cos 2A.\frac{\sin2^n A}{\sin 2^2A}

    Proceeding similarly, we find the R.H.S.

  57. We have to prove that 3(sin⁡A−cos⁡A)4+6(sin⁡A+cos⁡A)2+4(sin⁡6A+cos⁡6A)=133(\sin A - \cos A)^4 + 6(\sin A + \cos A)^2 + 4(\sin^6A + \cos^6A) = 13

    3(sin⁡A−cos⁡A)4=3[(sin⁡A−cos⁡A)2]2=3(1−sin⁡2A)23(\sin A - \cos A)^4 = 3[(\sin A - \cos A)^2]^2 = 3(1 - \sin 2A)^2

    6(sin⁡A+cos⁡A)2=6(1+sin⁡2A)6(\sin A + \cos A)^2 = 6(1 + \sin2A)

    4(sin⁡6A+cos⁡6A)=4[(cos⁡2A+sin⁡2A)3−3cos⁡2Asin⁡2A(cos⁡2A+sin⁡2A)]=4(1−34sin⁡22A)4(\sin^6A + \cos^6A) = 4[(\cos^2A + \sin^2A)^3 - 3\cos^2A\sin^2A(\cos^2A + \sin^2A)] = 4(1 - \frac{3}{4}\sin^22A)

    Adding all these yields 13.13.

  58. We have to prove that 2(sin⁡6A+cos⁡6A)−3(sin⁡4A+cos⁡4A)+1=02(\sin^6A + \cos^6A) - 3(\sin^4A + \cos^4A) + 1 = 0

    L.H.S. =2[(sin⁡2A+cos⁡2A)3−3sin⁡2Acos⁡2A(sin⁡2A+cos⁡2A)]−3[(sin⁡2A+cos⁡2A)2−2sin⁡2Acos⁡2A]+1= 2[(\sin^2A + \cos^2A)^3 - 3\sin^2A\cos^2A(\sin^2A + \cos^2A)] -3[(\sin^2A + \cos^2A)^2 - 2\sin^2A\cos^2A] + 1

    =2(1−3sin⁡2Acos⁡2A)−3[1−2sin⁡2Acos⁡2A]+1=0== 2(1 - 3\sin^2A\cos^2A) - 3[1 - 2\sin^2A\cos^2A] + 1 = 0 = R.H.S.

  59. Given cos⁡2A+cos⁡2(A+B)−2cos⁡Acos⁡Bcos⁡(A+B)\cos^2A + \cos^2(A + B) -2\cos A\cos B\cos(A + B)

    =cos⁡2A+cos⁡2(A+B)−2cos⁡Acos⁡Bcos⁡(A+B)+cos⁡2Acos⁡2B−cos⁡2Acos⁡2B= \cos^2A + \cos^2(A + B) -2\cos A\cos B\cos(A + B) + \cos^2A\cos^2B - \cos^2A\cos^2B

    =cos⁡2A+[cos⁡(A+B)−cos⁡Acos⁡B]2−cos⁡2Acos⁡2B= \cos^2A + [\cos(A + B) - \cos A\cos B]^2 - \cos^2A\cos^2B

    =cos⁡2A+sin⁡2Asin⁡2B−cos⁡2Acos⁡2B= \cos^2A + \sin^2A\sin^2B - \cos^2A\cos^2B

    =cos⁡2A+(1−cos⁡2A)(1−cos⁡2B)−cos⁡2Acos⁡2B= \cos^2A + (1 - \cos^2A)(1 - \cos^2B) - \cos^2A\cos^2B

    =1−cos⁡2B= 1 - \cos^2B which is independent of AA

  60. We have to prove that cos⁡3Acos⁡3A+sin⁡3Asin⁡3A=cos⁡32A\cos^3A\cos 3A + \sin^3A\sin 3A = \cos^32A

    We know that cos⁡3A=14(3cos⁡A+cos⁡3A)\cos^3A = \frac{1}{4}(3\cos A + \cos 3A) and

    sin⁡3A=14(3sin⁡A−sin⁡3A)\sin^3A = \frac{1}{4}(3\sin A - \sin 3A)

    L.H.S. =14(3cos⁡A+cos⁡3A)cos⁡3A+14(3sin⁡A−sin⁡3A)sin⁡3A= \frac{1}{4}(3\cos A + \cos 3A)\cos 3A + \frac{1}{4}(3\sin A - \sin 3A)\sin 3A

    =34(cos⁡3Acos⁡A+sin⁡3Asin⁡A)+14(cos⁡23A−sin⁡23A)= \frac{3}{4}(\cos3A\cos A + \sin 3A\sin A) + \frac{1}{4}(\cos^23A - \sin^23A)

    =34cos⁡2A+14cos⁡6A= \frac{3}{4}\cos 2A + \frac{1}{4}\cos6A

    =34cos⁡2A+14(4cos⁡32A−3cos⁡2A)= \frac{3}{4}\cos 2A + \frac{1}{4}(4\cos^32A - 3\cos 2A)

    =cos⁡32A== \cos^32A = R.H.S.

  61. We have to prove that tan⁡Atan⁡(60∘−A)tan⁡(60∘+A)=tan⁡3A\tan A\tan(60^\circ - A)\tan(60^\circ + A) = \tan 3A

    L.H.S. =sin⁡A.sin⁡(60∘−A).sin⁡(60∘+A)cos⁡A.cos⁡(60∘−A).cos⁡(60∘+A)= \frac{\sin A.\sin(60^\circ - A).\sin(60^\circ + A)}{\cos A.\cos(60^\circ - A).\cos(60^\circ + A)}

    =sin⁡A(sin⁡260∘−sin⁡2A)cos⁡A(cos⁡260∘−sin⁡2A)[∵sin⁡(A+B)sin⁡(A−B)=sin⁡2A−sin⁡2B and cos⁡(A+B)cos⁡(A−B)=cos⁡2A−sin⁡2B]= \frac{\sin A(\sin^260^\circ - \sin^2A)}{\cos A(\cos^260^\circ - \sin^2A)}[\because \sin(A + B)\sin (A - B) = \sin^2A - \sin^2B\text{~and~}\cos(A + B)\cos(A - B) = \cos^2A - \sin^2B]

    =sin⁡A(3−4sin⁡2A)cos⁡A(1−4sin⁡2A)=3sin⁡A−4sin⁡3A4cos⁡3A−3cos⁡A= \frac{\sin A(3 - 4\sin^2A)}{\cos A(1 - 4\sin^2A)} = \frac{3\sin A - 4\sin^3A}{4\cos^3A - 3\cos A}

    =sin⁡3Acos⁡3A=tan⁡3A== \frac{\sin 3A}{\cos 3A} = \tan 3A = R.H.S.

  62. We have to prove that sin⁡2A+sin⁡3(2π3+A)+sin⁡3(4π3+A)=−34sin⁡3A\sin^2A + \sin^3\left(\frac{2\pi}{3} + A\right) + \sin^3\left(\frac{4\pi}{3} + A\right) = -\frac{3}{4}\sin 3A

    ∵sin⁡3A=14[3sin⁡A−sin⁡3A]\because \sin^3A = \frac{1}{4}[3\sin A - \sin 3A]

    L.H.S. =14[3sin⁡A−sin⁡3A]+14[3sin⁡(2π3+A)−sin⁡(2π+3A)]+14[3sin⁡(4π3+A)−sin⁡(4π+3A)]= \frac{1}{4}[3\sin A - \sin 3A] + \frac{1}{4}\left[3\sin\left(\frac{2\pi}{3} + A\right) - \sin(2\pi + 3A)\right] + \frac{1}{4}\left[3\sin\left(\frac{4\pi}{3} + A\right) - \sin(4\pi + 3A)\right]

    =14[3sin⁡A−sin⁡3A]+14[3sin⁡(2π3+A)−sin⁡3A]+14[3sin⁡(4π3+A)−sin⁡3A]= \frac{1}{4}[3\sin A - \sin 3A] + \frac{1}{4}\left[3\sin\left(\frac{2\pi}{3} + A\right) - \sin3A\right] + \frac{1}{4}\left[3\sin\left(\frac{4\pi}{3} + A\right) - \sin3A\right]

    =34[sin⁡A−sin⁡3A+sin⁡(2π3+A)+sin⁡(4π3+A)]= \frac{3}{4}\left[\sin A - \sin 3A + \sin\left(\frac{2\pi}{3} + A\right) + \sin\left(\frac{4\pi}{3} + A\right)\right]

    =34[sin⁡A−sin⁡3A+2.(−sin⁡A).12]= \frac{3}{4}[\sin A - \sin 3A + 2.(-\sin A).\frac{1}{2}]

    =−34sin⁡3A== -\frac{3}{4}\sin 3A = R.H.S.

  63. We have to prove that 4(cos⁡310∘+sin⁡320∘)=3(cos⁡10∘+sin⁡20∘)4(\cos^310^\circ + \sin^320^\circ) = 3(\cos 10\circ + \sin 20^\circ)

    ⇒4cos⁡310∘−3cos⁡10∘=3sin⁡20∘−4sin⁡320∘\Rightarrow 4\cos^310^\circ - 3\cos10^\circ = 3\sin20^\circ - 4\sin^320^\circ

    ⇒cos⁡3.10∘=sin⁡3.20∘\Rightarrow \cos 3.10^\circ = \sin 3.20^\circ

    ⇒cos⁡30∘=sin⁡60∘⇒32=32\Rightarrow \cos 30^\circ = \sin 60^\circ \Rightarrow \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}

    Hence, proved.

  64. We have to prove that sin⁡Acos⁡3A−cos⁡Asin⁡3A=14sin⁡4A\sin A\cos^3A - \cos A\sin^3A = \frac{1}{4}\sin 4A

    L.H.S. =122sin⁡Acos⁡A(cos⁡2A−sin⁡2A)=12sin⁡2Acos⁡2A= \frac{1}{2}2\sin A\cos A(\cos^2A - \sin^2A) = \frac{1}{2}\sin2A\cos 2A

    =14.2.sin⁡2Acos⁡2A=14sin⁡4A== \frac{1}{4}.2.\sin2A\cos 2A = \frac{1}{4}\sin 4A = R.H.S.

  65. We have to prove that cos⁡3Asin⁡3A+sin⁡3Acos⁡3A=34sin⁡4A\cos^3A\sin3A + \sin^3A\cos 3A = \frac{3}{4}\sin 4A

    L.H.S. =cos⁡3A(3sin⁡A−4sin⁡3A)+sin⁡3A(4cos⁡3A−4cos⁡A)= \cos^3A(3\sin A - 4\sin^3A) + \sin^3A(4\cos^3A - 4\cos A)

    =3(sin⁡Acos⁡3A−cos⁡Asin⁡3A)= 3(\sin A\cos^3A - \cos A\sin^3A)

    Following previous problem we obtain R.H.S.

  66. We have to prove that sin⁡Asin⁡(60∘+A)sin⁡(A+120∘)=sin⁡3A\sin A\sin(60^\circ + A)\sin(A + 120^\circ) = \sin 3A

    We have proved in problem 32 that sin⁡Asin⁡(60∘−A)sin⁡(60∘+A)=14sin⁡3A\sin A\sin(60^\circ - A)\sin(60^\circ + A) = \frac{1}{4}\sin 3A

    Thus, we can prove what is required.

  67. We have to prove that cot⁡A+cot⁡(60∘+A)+cot⁡(120∘+A)=3cot⁡3A\cot A + \cot(60^\circ + A) + \cot(120^\circ + A) = 3\cot 3A

    L.H.S. =cot⁡A+cot⁡(60∘+A)+cot⁡(180∘−(60∘−A))= \cot A + \cot(60^\circ + A) + \cot(180^\circ - (60^\circ - A))

    =cot⁡A+cot⁡(60∘+A)−cot⁡(60∘−A)= \cot A + \cot(60^\circ + A) - \cot(60^\circ - A)

    This we have proved in problem 34.

  68. We have to prove that cos⁡5A=16cos⁡5A−20cos⁡3A+5cos⁡A\cos 5A = 16\cos^5A - 20\cos^3A + 5\cos A

    L.H.S. cos⁡(2A+3A)=cos⁡2Acos⁡3A−sin⁡2Asin⁡3A\cos(2A + 3A) = \cos 2A\cos 3A - \sin2A\sin3A

    =(2cos⁡2A−1)(4cos⁡3A−3cos⁡A)−2sin⁡Acos⁡A(3sin⁡A−4sin⁡3A)= (2\cos^2A - 1)(4\cos^3A - 3\cos A) - 2\sin A\cos A(3\sin A - 4\sin^3A)

    =8cos⁡5A−10cos⁡3A+3cos⁡A−2cos⁡Asin⁡2A[3−4(1−cos⁡2A)]= 8\cos^5A - 10\cos^3A + 3\cos A - 2\cos A\sin^2A[3 - 4(1 - \cos^2A)]

    =8cos⁡5A−10cos⁡3A+3cos⁡A−2cos⁡A(1−sin⁡2A)[4cos⁡2A−1]= 8\cos^5A - 10\cos^3A + 3\cos A - 2\cos A(1 - \sin^2A)[4\cos^2A - 1]

    =16cos⁡5A−20cos⁡3A+5cos⁡A== 16\cos^5A - 20\cos^3A + 5\cos A = R.H.S.

  69. We have to prove that sin⁡5A=5sin⁡A−20sin⁡3A+16sin⁡5A\sin 5A = 5\sin A - 20\sin^3A + 16\sin^5A

    L.H.S. =sin⁡5A=sin⁡(2A+3A)=sin⁡2Acos⁡3A+sin⁡3Acos⁡2A= \sin5A = \sin(2A + 3A) = \sin2A\cos3A + \sin3A\cos2A

    =2sin⁡Acos⁡A(4cos⁡3A−3cos⁡A)+(3sin⁡A−4sin⁡3A)(1−2sin⁡2A)= 2\sin A\cos A(4\cos^3A - 3\cos A) + (3\sin A - 4\sin^3A)(1 - 2\sin^2A)

    =2sin⁡A(1−sin⁡2A)(4cos⁡2A−3)+(3sin⁡A−4sin⁡3A)(1−2sin⁡2A)= 2\sin A(1 - \sin^2A)(4\cos^2A - 3) + (3\sin A - 4\sin^3A)(1 - 2\sin^2A)

    =2(sin⁡A−sin⁡3A)(1−4sin⁡2A)+(3sin⁡A−4sin⁡3A)(1−2sin⁡2A)= 2(\sin A - \sin^3A)(1 - 4\sin^2A) + (3\sin A - 4\sin^3A)(1 - 2\sin^2A)

    =5sin⁡A−20sin⁡3A+16sin⁡5A== 5\sin A - 20\sin^3A + 16\sin^5A = R.H.S.

  70. We have to prove that cos⁡4A−cos⁡4B=8(cos⁡A−cos⁡B)(cos⁡A+cos⁡B)(cos⁡A−sin⁡B)(cos⁡A+sin⁡B)\cos 4A - \cos 4B = 8(\cos A - \cos B)(\cos A + \cos B)(\cos A - \sin B)(\cos A + \sin B)

    R.H.S. =2(2cos⁡2A−2cos⁡2B)(2cos⁡2A−2sin⁡2B)= 2(2\cos^2A - 2\cos^2B)(2\cos^2A - 2\sin^2B)

    =2(cos⁡2A−cos⁡2B)(cos⁡2A+cos⁡2B)= 2(\cos 2A - \cos 2B)(\cos 2A + \cos 2B)

    =2(cos⁡22A−cos⁡22B)=cos⁡4A−cos⁡4B== 2(\cos^22A - \cos^22B) = \cos4A - \cos4B = L.H.S.

  71. We have to prove that tan⁡4A=4tan⁡A−4tan⁡3A1−6tan⁡2A+tan⁡4A\tan 4A = \frac{4\tan A - 4\tan^3A}{1 - 6\tan^2A + \tan^4A}

    L.H.S. =tan⁡4A=tan⁡(2A+2A)=2tan⁡2A1−tan⁡22A= \tan 4A = \tan(2A + 2A) = \frac{2\tan2A}{1 - \tan^22A}

    =2.2tan⁡A1−tan⁡2A1−(2tan⁡A1−tan⁡2A)2= \frac{2.\frac{2\tan A}{1 - \tan^2A}}{1 - \left(\frac{2\tan A}{1 - \tan^2A}\right)^2}

    Solving this yields R.H.S.

  72. Given 2tan⁡A=3tan⁡B,2\tan A = 3\tan B, we have to prove that tan⁡(A−B)=sin⁡2B5−cos⁡2B\tan (A- B) = \frac{\sin 2B}{5 - \cos 2B}

    tan⁡A=32tan⁡B\tan A = \frac{3}{2}\tan B

    tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B=32tan⁡B−tan⁡B1+32tan⁡2B\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A\tan B} = \frac{\frac{3}{2}\tan B - \tan B}{1 + \frac{3}{2}\tan^2B}

    =tan⁡B2+3tan⁡2B=sin⁡Bcos⁡B2cos⁡2B+3sin⁡2B= \frac{\tan B}{2 + 3\tan^2B} = \frac{\sin B\cos B}{2\cos^2B + 3\sin^2B}

    =sin⁡Bcos⁡B1+cos⁡2B+3.12(1−cos⁡2B)= \frac{\sin B\cos B}{1 + \cos 2B + 3.\frac{1}{2}(1 - \cos 2B)}

    =sin⁡2B5−cos⁡2B== \frac{\sin 2B}{5 - \cos 2B} = R.H.S.

  73. Given sin⁡A+sin⁡B=x\sin A + \sin B = x and cos⁡A+cos⁡B=y,\cos A + \cos B = y, we have to show that sin⁡(A+B)=2xyx2+y2\sin(A + B) = \frac{2xy}{x^2 + y^2}

    2xy=2(sin⁡A+sin⁡B)(cos⁡A+cos⁡B)2xy = 2(\sin A + \sin B)(\cos A + \cos B)

    =2(sin⁡Acos⁡A+sin⁡Bcos⁡B+sin⁡Acos⁡B+cos⁡Asin⁡B)= 2(\sin A\cos A + \sin B\cos B + \sin A\cos B + \cos A\sin B)

    =sin⁡2A+sin⁡2B+2sin⁡(A+B)= \sin2A + \sin 2B + 2\sin(A + B)

    =2sin⁡(A+B)cos⁡(A−B)+2sin⁡(A+B)=2sin⁡(A+B)[cos⁡(A−B)+1]= 2\sin(A + B)\cos(A - B) + 2\sin(A + B) = 2\sin(A + B)[\cos(A - B) + 1]

    x2+y2=(sin⁡A+sin⁡B)2+(cos⁡A+cos⁡B)2x^2 + y^2 = (\sin A + \sin B)^2 + (\cos A + \cos B)^2

    =2+2(cos⁡Acos⁡B+sin⁡Asin⁡B)=2[1+cos⁡(A−B)]= 2 + 2(\cos A\cos B + \sin A \sin B) = 2[1 + \cos (A - B)]

    ∴2xyx2+y2=sin⁡(A+B)\therefore \frac{2xy}{x^2 + y^2} = \sin(A + B)

  74. Given A=π2n+1,A= \frac{\pi}{2^n + 1}, we have to prove that cos⁡A.cos⁡2A.cos⁡22A.….cos⁡2n−1A=12n\cos A.\cos 2A. \cos2^2A.\ldots.\cos2^{n - 1}A = \frac{1}{2^n}

    L.H.S. =cos⁡A.cos⁡2A.cos⁡22A.….cos⁡2n−1A= \cos A.\cos 2A. \cos2^2A.\ldots.\cos2^{n - 1}A

    =12sin⁡A(2sin⁡Acos⁡A).cos⁡2A.cos⁡22A.….cos⁡2n−1A= \frac{1}{2\sin A}(2\sin A\cos A).\cos 2A. \cos2^2A.\ldots.\cos2^{n - 1}A

    =12sin⁡Asin⁡2A.cos⁡2A.cos⁡22A.….cos⁡2n−1A= \frac{1}{2\sin A}\sin 2A.\cos 2A.\cos2^2A.\ldots.\cos2^{n - 1}A

    =122sin⁡A(2sin⁡2Acos⁡2A)cos⁡22A.….cos⁡2n−1A= \frac{1}{2^2\sin A}(2\sin 2A\cos 2A)\cos2^2A.\ldots.\cos2^{n - 1}A

    Proceeding similarly

    =12nsin⁡Asin⁡2nA=12nsin⁡Asin⁡(π−A)=12n== \frac{1}{2^n\sin A}\sin2^n A = \frac{1}{2^n\sin A}\sin(\pi - A) = \frac{1}{2^n} = R.H.S.

  75. Given tan⁡A=yx,\tan A = \frac{y}{x}, we have to prove that xcos⁡2A+ysin⁡2A=xx\cos 2A + y\sin 2A = x

    ∵tan⁡A=yx∴sin⁡A=yx2+y2,cos⁡A=xx2+y2\because \tan A = \frac{y}{x} \therefore \sin A = \frac{y}{\sqrt{x^2 + y^2}}, \cos A = \frac{x}{\sqrt{x^2 + y^2}}

    ∴xcos⁡2A+ysin⁡2A=x(cos⁡2A−sin⁡2A)+2ysin⁡Acos⁡A=x(x2−y2x2+y2)+2x2yx2+y2\therefore x\cos 2A + y\sin 2A = x(\cos^2A - \sin^2A) + 2y\sin A\cos A = x\left(\frac{x^2 - y^2}{x^2 + y^2}\right) + 2\frac{x^2y}{x^2 + y^2}

    =x== x = R.H.S.

  76. Given tan⁡2A=1+2tan⁡2B,\tan^2A = 1 + 2\tan^2B, we have to prove that cos⁡2B=1+2cos⁡2A\cos 2B = 1 + 2\cos 2A

    1+2cos⁡2A=1+2.1−tan⁡2A1+tan⁡2A=3−tan⁡2A1+tan⁡2A1 + 2\cos 2A = 1 + 2.\frac{1 - \tan^2A}{1 + \tan^2A} = \frac{3 - \tan^2A}{1 + \tan^2A}

    =3−1−2tan⁡2B2+2tan⁡2B=1−tan⁡2B1+tan⁡2B=cos⁡2B== \frac{3 - 1 - 2\tan^2B}{2 + 2\tan^2B} = \frac{1 - \tan^2B}{1 + \tan^2B} = \cos 2B = L.H.S.

  77. Given cos⁡2A=3cos⁡2B−13−cos⁡2B,\cos 2A = \frac{3\cos 2B - 1}{3 - \cos 2B}, we have to prove that tan⁡A=2tan⁡B\tan A = \sqrt{2}\tan B

    cos⁡2A=3cos⁡2B−13−cos⁡2B\cos 2A = \frac{3\cos 2B - 1}{3 - \cos 2B}

    ⇒1−tan⁡2A1+tan⁡B=3−3tan⁡2B−1−tan⁡2B3+3tan⁡2B−1+tan⁡2B\Rightarrow \frac{1 - \tan^2A}{1 + \tan^B} = \frac{3 - 3\tan^2B - 1 - \tan^2B}{3 + 3\tan^2B - 1 + \tan^2B}

    =1−2tan⁡2B1+2tan⁡2B= \frac{1 - 2\tan^2B}{1 + 2\tan^2B}

    ∴tan⁡2A=2tan⁡2B⇒tan⁡A=2tan⁡B\therefore \tan^2A = 2\tan^2B \Rightarrow \tan A = \sqrt{2}\tan B

  78. Given tan⁡B=3tan⁡A,\tan B = 3\tan A, we have to prove that tan⁡(A+B)=2sin⁡2B1+cos⁡2B\tan(A + B) = \frac{2\sin 2B}{1 + \cos 2B}

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

    =43tan⁡B1−tan⁡2B3=4tan⁡B3−tan⁡2B= \frac{\frac{4}{3}\tan B}{1 - \frac{\tan^2B}{3}} = \frac{4\tan B}{3 - \tan^2B}

    =4sin⁡Bcos⁡B3cos⁡2B−sin⁡2B=2sin⁡2B2cos⁡2B+cos⁡2B= \frac{4\sin B\cos B}{3\cos^2B - \sin^2B} = \frac{2\sin2B}{2\cos^2B + \cos2B}

    =2sin⁡2B1+cos⁡2B== \frac{2\sin2B}{1 + \cos2B} = R.H.S.

  79. Given xsin⁡A=ycos⁡A,x\sin A = y\cos A, we have to prove that xsec⁡2A+ycosec⁡2A=x\frac{x}{\sec 2A} + \frac{y}{\cosec 2A} = x

    Given tan⁡A=yx∴sin⁡A=yx2+y2&cos⁡A=xx2+y2\tan A = \frac{y}{x} \therefore \sin A = \frac{y}{\sqrt{x^2 + y^2}} \& \cos A = \frac{x}{\sqrt{x^2 + y^2}}

    L.H.S. =xsec⁡2A+ycosec⁡2A= \frac{x}{\sec 2A} + \frac{y}{\cosec 2A}

    =xcos⁡2A+ysin⁡2A=x(cos⁡2A−sin⁡2A)+2ysin⁡Acos⁡A= x\cos2A + y\sin2A = x(\cos^2A - \sin^2A) + 2y\sin A\cos A

    xx2−y2x2+y2+2xy2x2+y2x\frac{x^2 - y^2}{x^2 + y^2} + \frac{2xy^2}{x^2 + y^2}

    =x= x

  80. Given tan⁡A=sec⁡2B,\tan A = \sec 2B, we have to prove that sin⁡2A=1−tan⁡4B1+tan⁡4B\sin 2A = \frac{1 - \tan^4B}{1 + \tan^4B}

    tan⁡A=1cos⁡2B=1+tan⁡2B1−tan⁡2B\tan A = \frac{1}{\cos 2B} = \frac{1 + \tan^2B}{1 - \tan^2B}

    ∴sin⁡A=1+tan⁡2B2+2tan⁡4B\therefore \sin A = \frac{1 + \tan^2B}{\sqrt{2 + 2\tan^4B}}

    and cos⁡A=1−tan⁡2B2+2tan⁡4B\cos A = \frac{1 - \tan^2B}{\sqrt{2 + 2\tan^4B}}

    L.H.S. sin⁡2A=2sin⁡Acos⁡A=1−tan⁡4B1+tan⁡4B=\sin 2A = 2\sin A\cos A = \frac{1 - \tan^4B}{1 + \tan^4B} = R.H.S.

  81. Given A=π3,A = \frac{\pi}{3}, we have to prove that cos⁡A.cos⁡2A.cos⁡3A.cos⁡4A.cos⁡5A.cos⁡6A=−116\cos A.\cos 2A. \cos 3A.\cos 4A.\cos 5A.\cos 6A = -\frac{1}{16}

    L.H.S. =182cos⁡A.cos⁡6A.2cos⁡2A.cos⁡5A.2cos⁡3Acos⁡4A= \frac{1}{8}2\cos A.\cos 6A.2\cos 2A.\cos 5A.2\cos 3A\cos 4A

    =18(cos⁡7A2+cos⁡5A2)(cos⁡7A2+cos⁡3A2)(cos⁡7A2+cos⁡A2)= \frac{1}{8}\left(\cos \frac{7A}{2} + \cos \frac{5A}{2}\right)\left(\cos \frac{7A}{2} + \cos \frac{3A}{2}\right)\left(\cos \frac{7A}{2} + \cos \frac{A}{2}\right)

    =18[cos⁡(π+π6)+cos⁡(π−π6)][cos⁡(π+π6)+cos⁡(π2)]+[cos⁡(π+π6)+cos⁡π6]= \frac{1}{8}\left[\cos\left(\pi + \frac{\pi}{6}\right) + \cos \left(\pi - \frac{\pi}{6}\right)\right]\left[\cos\left(\pi + \frac{\pi}{6}\right) + \cos \left(\frac{\pi}{2}\right)\right] + \left[\cos\left(\pi + \frac{\pi}{6}\right) + \cos \frac{\pi}{6}\right]

    =−116= -\frac{1}{16}

  82. Given A=π15,A = \frac{\pi}{15}, we have to prove that cos⁡2A.cos⁡4A.cos⁡8A.cos⁡14A=116\cos2A.\cos4A.\cos8A.\cos14A = \frac{1}{16}

    cos⁡14A=cos⁡14π15=cos⁡(2π−16π15)=cos⁡16A\cos 14A = \cos \frac{14\pi}{15} = \cos \left(2\pi - \frac{16\pi}{15}\right) = \cos 16A

    L.H.S. =cos⁡2A.cos⁡4A.cos⁡8A.cos⁡16A=12sin⁡2A.2sin⁡2A.cos⁡2A.cos⁡4A.cos⁡8A.cos⁡16A= \cos2A.\cos4A.\cos8A.\cos16A = \frac{1}{2\sin2A}.2\sin2A.\cos2A.\cos4A.\cos8A.\cos16A

    =12sin⁡2Asin⁡4A.cos⁡4A.cos⁡8A.cos⁡16A=122sin⁡2Asin⁡8A.cos⁡8A.cos⁡16A= \frac{1}{2\sin2A}\sin4A.\cos4A.\cos8A.\cos16A = \frac{1}{2^2\sin 2A}\sin8A.\cos8A.\cos16A

    =124sin⁡2Asin⁡32A=116sin⁡2Asin⁡(2π+2A)=116== \frac{1}{2^4\sin 2A}\sin32A = \frac{1}{16\sin2A}\sin(2\pi + 2A) = \frac{1}{16} = R.H.S.

  83. Given tan⁡Atan⁡B=a−ba+b,\tan A\tan B = \sqrt{\frac{a - b}{a + b}}, we have to prove that (a−bcos⁡2A)(a−bcos⁡2B)=a2−b2(a - b\cos2A)(a - b\cos2B) = a^2 - b^2

    L.H.S. =(a−bcos⁡2A)(a−bcos⁡2B)=[a−b1−tan⁡2A1+tan⁡2A][a−b1−tan⁡2B1+tan⁡2B]= (a - b\cos2A)(a - b\cos2B) = \left[a -b\frac{1 - \tan^2A}{1 + \tan^2A}\right]\left[a - b\frac{1 - \tan^2B}{1 + \tan^2B}\right]

    =[a−b1−tan⁡2A1+tan⁡2A][a−b1−a−b(a+b)tan⁡2A1+a−b(a+b)tan⁡2A]= \left[a -b\frac{1 - \tan^2A}{1 + \tan^2A}\right]\left[a - b\frac{1 - \frac{a - b}{(a + b)\tan^2A}}{1 + \frac{a - b}{(a + b)\tan^2A}}\right]

    Solving this yields a2−b2\frac{a^2 - b^2}{}

  84. Given sin⁡A=12\sin A = \frac{1}{2} and sin⁡B=13,\sin B = \frac{1}{3}, we have to find the value of sin⁡(A+B)\sin(A + B) and sin⁡(2A+2B)\sin(2A + 2B)

    cos⁡A=32\cos A = \frac{\sqrt{3}}{2} and cos⁡B=83\cos B = \frac{\sqrt{8}}{3}

    sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B=86+36=8+36\sin(A + B) = \sin A\cos B + \cos A\sin B = \frac{\sqrt{8}}{6} + \frac{\sqrt{3}}{6} = \frac{\sqrt{8} + \sqrt{3}}{6}

    sin⁡(2A+2B)=sin⁡2Acos⁡2B+cos⁡2Asin⁡2B\sin(2A + 2B) = \sin 2A\cos 2B + \cos 2A\sin 2B

    =2sin⁡Acos⁡A(cos⁡2B−sin⁡2B)+2sin⁡Bcos⁡B(cos⁡2A−sin⁡2A)= 2\sin A\cos A(\cos^2B - \sin^2B) + 2\sin B\cos B(\cos^2A - \sin^2A)

    Substituting the values we obtain the desired result.

85 and 86 have been left as exercises.

  1. cos⁡A=310=1−tan⁡2A21+tan⁡2A2=310\cos A = \frac{3}{10} = \frac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}} = \frac{3}{10}

    Let x=tan⁡A2,x = \tan \frac{A}{2}, then 1−x21+x2=310\frac{1 - x^2}{1 + x^2} = \frac{3}{10}

    x=±713x = \pm \sqrt{\frac{7}{13}}

    The reason for two values is that cos⁡A\cos A may lie in first or fourth quadrant. If it is in first quadrant then tan⁡A2\tan \frac{A}{2} will be positive and if it is in fourth quadrant then tan⁡A2\tan \frac{A}{2} will be negative.

  2. Given sin⁡A+sin⁡B=x\sin A + \sin B = x and cos⁡A+cos⁡B=y,\cos A + \cos B = y, we have to find the value of tan⁡A−B2\tan \frac{A - B}{2}

    tan⁡A−B2=tan⁡A2−tan⁡B21+tan⁡A2tan⁡B2\tan \frac{A - B}{2} = \frac{\tan \frac{A}{2} - \tan \frac{B}{2}}{1 + \tan\frac{A}{2}\tan\frac{B}{2}}

    =sin⁡A2cos⁡B2−sin⁡B2cos⁡A2cos⁡A2cos⁡B2−sin⁡A2sin⁡B2= \frac{\sin \frac{A}{2}\cos\frac{B}{2} - \sin\frac{B}{2}\cos\frac{A}{2}}{\cos\frac{A}{2}\cos\frac{B}{2} - \sin\frac{A}{2}\sin\frac{B}{2}}

    Also, tan⁡(A−B)=2tan⁡A−B21−tan⁡2A−B2\tan(A - B) = \frac{2\tan\frac{A - B}{2}}{1 - \tan^2\frac{A - B}{2}}

    Let tan⁡A−B2=a,\tan\frac{A - B}{2} = a, then tan⁡(A−B)=2a1+a2\tan(A - B) = \frac{2a}{1 + a^2}

    x2+y2=2+2sin⁡Asin⁡B+2cos⁡Acos⁡Bx^2 + y^2 = 2 + 2\sin A\sin B + 2\cos A\cos B

    Solving this yields tan⁡A−B2=4−x2−y2x2+y2\tan\frac{A - B}{2} = \sqrt{\frac{4 - x^2 - y^2}{x^2 + y^2}}

  3. We have to prove that (cos⁡A+cos⁡B)2+(sin⁡A−sin⁡B)2=4cos⁡2A+B2(\cos A + \cos B)^2 + (\sin A - \sin B)^2 = 4\cos^2 \frac{A + B}{2}

    L.H.S. =cos⁡2A+cos⁡2B+2cos⁡Acos⁡B+sin⁡2A+sin⁡2B−2sin⁡Asin⁡B= \cos^2A + \cos^2B + 2\cos A\cos B + \sin^2A + \sin^2B - 2\sin A\sin B

    =2+2cos⁡(A+B)=4cos⁡2A+B2== 2 + 2\cos(A + B) = 4\cos^2\frac{A + B}{2} = R.H.S.

  4. We have to prove that (cos⁡A+cos⁡B)2+(sin⁡A+sin⁡B)2=4cos⁡2A−B2(\cos A + \cos B)^2 + (\sin A + \sin B)^2 = 4\cos^2 \frac{A - B}{2}

    L.H.S. =cos⁡2A+cos⁡2B+2cos⁡Acos⁡B+sin⁡2A+sin⁡2B+2sin⁡Asin⁡B= \cos^2A + \cos^2B + 2\cos A\cos B + \sin^2A + \sin^2B + 2\sin A\sin B

    =2+2cos⁡(A−B)=4cos⁡2A−B2== 2 + 2\cos(A - B) = 4\cos^2\frac{A - B}{2} = R.H.S.

  5. We hve to prove that (cos⁡A−cos⁡B)2+(sin⁡A−sin⁡B)2=4sin⁡2A−B2(\cos A - \cos B)^2 + (\sin A - \sin B)^2 = 4\sin^2 \frac{A - B}{2}

    L.H.S. =cos⁡2A+cos⁡2B−2cos⁡Acos⁡B+sin⁡2A+sin⁡2B+2sin⁡Asin⁡B= \cos^2A + \cos^2B - 2\cos A\cos B + \sin^2A + \sin^2B + 2\sin A\sin B

    =2−2cos⁡(A−B)=4sin⁡2A−B2== 2 - 2\cos(A - B) = 4\sin^2\frac{A - B}{2} = R.H.S.

  6. We have to prove that sin⁡2(π8+A2)−sin⁡2(π8−A2)=12sin⁡A\sin^2\left(\frac{\pi}{8} + \frac{A}{2}\right) - \sin^2\left(\frac{\pi}{8} -\frac{A}{2}\right) = \frac{1}{\sqrt{2}}\sin A

    L.H.S. =1−cos⁡(π4+A)2−1−cos⁡(π4−A)2= \frac{1 - \cos\left(\frac{\pi}{4} + A\right)}{2} - \frac{1 - \cos\left(\frac{\pi}{4} - A\right)}{2}

    =cos⁡(π4−A)−cos⁡(π4+A)2= \frac{\cos\left(\frac{\pi}{4} - A\right) - \cos\left(\frac{\pi}{4} + A\right)}{2}

    =2sin⁡π4sin⁡A2=12sin⁡A== \frac{2\sin\frac{\pi}{4}\sin A}{2} = \frac{1}{\sqrt{2}}\sin A = R.H.S.

  7. We have to prove that (tan⁡4A+tan⁡2A)(1−tan⁡23Atan⁡2A)=2tan⁡3Asec⁡2A(\tan 4A + \tan 2A)(1 - \tan^23A\tan^2A) = 2\tan 3A\sec^2A

    L.H.S. =(tan⁡4A+tan⁡2A)(1+tan⁡3Atan⁡A)(1−tan⁡3Atan⁡A)= (\tan 4A + \tan 2A)(1 + \tan 3A\tan A)(1 - \tan 3A\tan A)

    =(sin⁡4Acos⁡4A+sin⁡2Acos⁡2A)(cos3Acos⁡A+sin⁡3Asin⁡Acos⁡3Acos⁡A)(cos3Acos⁡A−sin⁡3Asin⁡Acos⁡3A−cos⁡A)= \left(\frac{\sin 4A}{\cos 4A} + \frac{\sin 2A}{\cos 2A}\right)\left(\frac{cos 3A\cos A + \sin 3A\sin A}{\cos 3A\cos A}\right)\left(\frac{cos 3A\cos A - \sin 3A\sin A}{\cos 3A - \cos A}\right)

    =sin⁡6Acos⁡4Acos⁡2A.cos⁡4Acos⁡3Acos⁡Acos⁡2Acos⁡3Acos⁡A= \frac{\sin 6A}{\cos 4A\cos 2A}.\frac{\cos 4A}{\cos 3A\cos A}\frac{\cos 2A}{\cos 3A\cos A}

    =2sin⁡3Acos⁡3Acos⁡23Acos⁡2A=2tan⁡3Asec⁡2A== \frac{2\sin3A\cos3A}{\cos^23A\cos^2A} = 2\tan3A\sec^2A = R.H.S.

  8. We have to prove that (1+tan⁡A2−sec⁡A2)(1+tan⁡A2+sec⁡A2)=sin⁡Asec⁡2A2\left(1 + \tan \frac{A}{2} - \sec\frac{A}{2}\right)\left(1 + \tan \frac{A}{2} + \sec\frac{A}{2}\right) = \sin A\sec^2\frac{A}{2}

    L.H.S. =(1+tan⁡A2−sec⁡A2)(1+tan⁡A2+sec⁡A2)= \left(1 + \tan \frac{A}{2} - \sec\frac{A}{2}\right)\left(1 + \tan \frac{A}{2} + \sec\frac{A}{2}\right)

    =(1+tan⁡A2)2−sec2A2=2tan⁡A2= \left(1 + \tan\frac{A}{2}\right)^2 - sec^2\frac{A}{2} = 2\tan\frac{A}{2}

    =2sin⁡A2cos⁡A2cos⁡2A2=sin⁡Asec⁡2A2== \frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{\cos^2\frac{A}{2}} = \sin A\sec^2\frac{A}{2} = R.H.S.

  9. We have to prove that 1+sin⁡A−cos⁡A1+sin⁡A+cos⁡A=tan⁡A2\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \frac{A}{2}

    L.H.S. =(1−cos⁡A)+sin⁡A(1+cos⁡A)+sin⁡A= \frac{(1 - \cos A) + \sin A}{(1 + \cos A) + \sin A}

    =2sin⁡2A2+2sin⁡A2cos⁡A22cos⁡2A2+2sin⁡A2cos⁡A2= \frac{2\sin^2\frac{A}{2} + 2\sin\frac{A}{2}\cos\frac{A}{2}}{2\cos^2\frac{A}{2} + 2\sin\frac{A}{2}\cos\frac{A}{2}}

    =sin⁡A2(sin⁡A2+cos⁡A2)cos⁡A2(sin⁡A2+cos⁡A2)= \frac{\sin\frac{A}{2}\left(\sin\frac{A}{2} + \cos\frac{A}{2}\right)}{\cos\frac{A}{2}\left(\sin\frac{A}{2} + \cos\frac{A}{2}\right)}

    =tan⁡A2== \tan\frac{A}{2} = R.H.S.

  10. We have to prove that 1−tan⁡A21+tan⁡A2=1+sin⁡Acos⁡A=tan⁡(π4+A2)\frac{1 - \tan \frac{A}{2}}{1 + \tan \frac{A}{2}} = \frac{1 + \sin A}{\cos A} = \tan \left(\frac{\pi}{4} + \frac{A}{2}\right)

    1+sin⁡Acos⁡A=sin⁡2A2+cos⁡2A2+2sin⁡A2cos⁡A2cos⁡2A2−sin⁡2A2\frac{1 + \sin A}{\cos A} = \frac{\sin^2\frac{A}{2} + \cos^2\frac{A}{2} + 2\sin\frac{A}{2}\cos \frac{A}{2}}{\cos^2\frac{A}{2} - \sin^2\frac{A}{2}}

    =sin⁡A2+cos⁡A2cos⁡A2−sin⁡A2= \frac{\sin\frac{A}{2} + \cos\frac{A}{2}}{\cos \frac{A}{2} - \sin \frac{A}{2}}

    Dividing numerator and denominator by cos⁡A2,\cos \frac{A}{2}, we get

    =1+tan⁡A21−tan⁡A2= \frac{1 + \tan \frac{A}{2}}{1 - \tan \frac{A}{2}}

  11. We have to prove that cos⁡4π8+cos⁡43π8+cos⁡45π8+cos⁡47π8=32\cos^4\frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4\frac{5\pi}{8} + \cos^4\frac{7\pi}{8}= \frac{3}{2}

    cos⁡4π8=(cos⁡2π8)2=(1+cos⁡π42)2\cos^4\frac{\pi}{8} = \left(\cos^2\frac{\pi}{8}\right)^2 = \left(\frac{1 + \cos\frac{\pi}{4}}{2}\right)^2

    =(1+122)2=38+24= \left(\frac{1 + \frac{1}{\sqrt{2}}}{2}\right)^2 = \frac{3}{8} + \frac{\sqrt{2}}{4}

    Similalry, cos⁡43π8=38−24\cos^4\frac{3\pi}{8} = \frac{3}{8} - \frac{\sqrt{2}}{4}

    cos⁡5π8=cos⁡(π−3π8)=−cos⁡3π8\cos\frac{5\pi}{8} = \cos\left(\pi - \frac{3\pi}{8}\right) = -\cos\frac{3\pi}{8}

    cos⁡7π8=−cos⁡π8\cos\frac{7\pi}{8} = -\cos\frac{\pi}{8}

    Thus, cos⁡4π8+cos⁡43π8+cos⁡45π8+cos⁡47π8=32\cos^4\frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4\frac{5\pi}{8} + \cos^4\frac{7\pi}{8}= \frac{3}{2}

  12. We have to prove that 2sin⁡A−sin⁡2A2sin⁡A+sin⁡2A=tan⁡2A2\frac{2\sin A - \sin2A}{2\sin A + \sin 2A} = \tan^2\frac{A}{2}

    L.H.S. =2sin⁡A−2sin⁡Acos⁡A2cos⁡A+2sin⁡Acos⁡A=2sin⁡A(1−cos⁡A)2sin⁡A(1+cos⁡A)= \frac{2\sin A - 2\sin A\cos A}{2\cos A + 2\sin A\cos A} = \frac{2\sin A(1 - \cos A)}{2\sin A(1 + \cos A)}

    =2sin⁡2A22cos⁡2A2=tan⁡2A2== \frac{2\sin^2\frac{A}{2}}{2\cos^2\frac{A}{2}} = \tan^2\frac{A}{2} = R.H.S.

  13. We have to prove that cot⁡A2−tan⁡A2=2cot⁡A\cot \frac{A}{2} - \tan \frac{A}{2} = 2\cot A

    L.H.S. =cos⁡A2sin⁡A2−sin⁡A2cos⁡A2= \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}} - \frac{\sin \frac{A}{2}}{\cos\frac{A}{2}}

    =cos⁡2A2−sin⁡2A2sin⁡A2cos⁡A2= \frac{\cos^2\frac{A}{2} - \sin^2\frac{A}{2}}{\sin \frac{A}{2}\cos\frac{A}{2}}

    =2cos⁡Asin⁡A=2cot⁡A== \frac{2\cos A}{\sin A} = 2\cot A = R.H.S.

  14. We have to prove that 1+sin⁡A1−sin⁡A=tan⁡2(π4+A2)\frac{1 + \sin A}{1 - \sin A} = \tan^2\left(\frac{\pi}{4} + \frac{A}{2}\right)

    L.H.S. =cos⁡2A2+sin⁡2A2+2cos⁡A2sin⁡A2cos⁡2A2+sin⁡2A2−2cos⁡A2sin⁡A2= \frac{\cos^2\frac{A}{2} + \sin^2\frac{A}{2} + 2\cos \frac{A}{2}\sin\frac{A}{2}}{\cos^2\frac{A}{2} + \sin^2\frac{A}{2} - 2\cos \frac{A}{2}\sin\frac{A}{2}}

    =cos⁡A2+sin⁡A2cos⁡A2−sin⁡A2= \frac{\cos \frac{A}{2} + \sin \frac{A}{2}}{\cos \frac{A}{2} - \sin \frac{A}{2}}

    Dividing both numerator and denominator by cos⁡A2,\cos \frac{A}{2}, we get

    =1+tan⁡A21−tan⁡A2=tan⁡π4+tan⁡A21−tan⁡π4tan⁡A2= \frac{1 + \tan\frac{A}{2}}{1 - \tan \frac{A}{2}} = \frac{\tan\frac{\pi}{4} + \tan \frac{A}{2}}{1 - \tan\frac{\pi}{4}\tan\frac{A}{2}}

    =tan⁡(π4+A2)== \tan\left(\frac{\pi}{4} + \frac{A}{2}\right) = R.H.S.

  15. We have to prove that sec⁡A+tan⁡A=tan⁡(π4+A2)\sec A + \tan A = \tan\left(\frac{\pi}{4} + \frac{A}{2}\right)

    L.H.S. =1+sin⁡Acos⁡A=(cos⁡A2+sin⁡A2)2cos⁡2A2−sin⁡2A2= \frac{1 + \sin A}{\cos A} = \frac{\left(\cos\frac{A}{2} + \sin\frac{A}{2}\right)^2}{\cos^2\frac{A}{2} - \sin^2\frac{A}{2}}

    =cos⁡A2+sin⁡A2cos⁡A2−sin⁡A2= \frac{\cos \frac{A}{2} + \sin \frac{A}{2}}{\cos \frac{A}{2} - \sin \frac{A}{2}}

    Now proceeding like previous problem

    =tan⁡(π4+A2)== \tan\left(\frac{\pi}{4} + \frac{A}{2}\right) = R.H.S.

  16. We have to prove that sin⁡A+sin⁡B−sin⁡(A+B)sin⁡A+sin⁡B+sin⁡(A+B)=tan⁡A2tan⁡B2\frac{\sin A + \sin B - \sin(A + B)}{\sin A + \sin B + \sin(A + B)} = \tan \frac{A}{2}\tan \frac{B}{2}

    L.H.S. =2sin⁡A+B2cos⁡A−B2−2sin⁡A+B2cos⁡A+B22sin⁡A+B2cos⁡A−B2+2sin⁡A+B2cos⁡A+B2= \frac{2\sin\frac{A + B}{2}\cos\frac{A - B}{2} - 2\sin \frac{A + B}{2}\cos\frac{A + B}{2}}{2\sin\frac{A + B}{2}\cos\frac{A - B}{2} + 2\sin \frac{A + B}{2}\cos\frac{A + B}{2}}

    =cos⁡A−B2−cos⁡A+B2cos⁡A−B2+cos⁡A+B2= \frac{\cos\frac{A - B}{2} - \cos \frac{A + B}{2}}{\cos\frac{A - B}{2} + \cos \frac{A + B}{2}}

    =2sin⁡A2cos⁡B22cos⁡A2cos⁡B2= \frac{2\sin\frac{A}{2}\cos\frac{B}{2}}{2\cos\frac{A}{2}\cos\frac{B}{2}}

    =tan⁡A2tan⁡B2== \tan \frac{A}{2}\tan \frac{B}{2} = R.H.S.

  17. We have to prove that tan⁡(π4−A2)=sec⁡A−tan⁡A=1−sin⁡A1+sin⁡A\tan \left(\frac{\pi}{4} - \frac{A}{2}\right) = \sec A - \tan A = \sqrt{\frac{1 - \sin A}{1 + \sin A}}

    L.H.S. =tan⁡(π4−A2)=1−tan⁡A21+tan⁡A2= \tan \left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{1 - \tan \frac{A}{2}}{1 + \tan \frac{A}{2}}

    =cos⁡A2−sin⁡A2cos⁡A2+sin⁡A2= \frac{\cos\frac{A}{2} - \sin \frac{A}{2}}{\cos\frac{A}{2} + \sin \frac{A}{2}}

    Multiplying both numerator and denominator by cos⁡A2+sin⁡A2\cos\frac{A}{2} + \sin\frac{A}{2}

    =cos⁡A1+sin⁡A=cos⁡2A(1+sin⁡A)2=1−sin⁡A1+sin⁡A= \frac{\cos A}{1 + \sin A} = \sqrt{\frac{\cos^2A}{(1 + \sin A)^2}} = \sqrt{\frac{1 - \sin A}{1 + \sin A}}

    Also, cos⁡A1+sin⁡A=cos⁡A(1−sin⁡A)1−sin⁡2A=sec⁡A−tan⁡A\frac{\cos A}{1 + \sin A} = \frac{\cos A(1 - \sin A)}{1 - \sin^2A} = \sec A - \tan A

  18. We have to prove that cosec⁡(π4+A2)cosec⁡(π4−A2)=2sec⁡A\cosec\left(\frac{\pi}{4} + \frac{A}{2}\right)\cosec \left(\frac{\pi}{4} - \frac{A}{2}\right) = 2\sec A

    L.H.S. =1sin⁡(π4+A2).1sin⁡(π4−A2)= \frac{1}{\sin\left(\frac{\pi}{4} + \frac{A}{2}\right)}.\frac{1}{\sin\left(\frac{\pi}{4} - \frac{A}{2}\right)}

    =2cos⁡A−cos⁡π2=2sec⁡A== \frac{2}{\cos A - \cos \frac{\pi}{2}} = 2\sec A = R.H.S.

  19. We have to prove that cos⁡2π8+cos⁡23π8+cos⁡25π8+cos⁡27π8=2\cos^2\frac{\pi}{8} + \cos^2\frac{3\pi}{8} + \cos^2\frac{5\pi}{8} + \cos^2\frac{7\pi}{8} = 2

    cos⁡2π8=1+cos⁡π42=1+222\cos^2\frac{\pi}{8} = \frac{1 + \cos \frac{\pi}{4}}{2} = \frac{1 + \sqrt{2}}{2\sqrt{2}}

    cos⁡23π8=1+cos⁡3π42=2−122\cos^2\frac{3\pi}{8} = \frac{1 + \cos \frac{3\pi}{4}}{2} = \frac{\sqrt{2} - 1}{2\sqrt{2}}

    cos⁡25π8=cos⁡23π8\cos^2\frac{5\pi}{8} = \cos^2\frac{3\pi}{8}

    cos⁡27π8=cos⁡2π8\cos^2\frac{7\pi}{8} = \cos^2\frac{\pi}{8}

    L.H.S. =2(1+222+2−122)= 2\left(\frac{1 + \sqrt{2}}{2\sqrt{2}} + \frac{\sqrt{2} - 1}{2\sqrt{2}}\right)

    =2== 2 = R.H.S.

  20. This problem is similar to previous problem and can be solved in a likewise manner.

  21. We have to prove that (1+cos⁡π8)(1+cos⁡3π8)(1+cos⁡5π8)(1+cos⁡7π8)=18\left(1 + \cos \frac{\pi}{8}\right)\left(1 + \cos\frac{3\pi}{8}\right)\left(1 + \cos\frac{5\pi}{8}\right)\left(1 + \cos \frac{7\pi}{8}\right) = \frac{1}{8}

    cos⁡7π8=cos⁡(π−π8)=−cos⁡π8\cos\frac{7\pi}{8} = \cos\left(\pi - \frac{\pi}{8}\right) = -\cos\frac{\pi}{8}

    cos⁡5π8=cos⁡(π−3π8)=−cos⁡3π8\cos\frac{5\pi}{8} = \cos\left(\pi - \frac{3\pi}{8}\right) = -\cos\frac{3\pi}{8}

    L.H.S. =(1−cos⁡2π8)(1−cos⁡23π8)= \left(1 - \cos^2\frac{\pi}{8}\right)\left(1 - \cos^2\frac{3\pi}{8}\right)

    =sin⁡2π8sin⁡23π8= \sin^2\frac{\pi}{8}\sin^2\frac{3\pi}{8}

    =1−2cos⁡π42.1−2cos⁡3π42= \frac{1 - 2\cos\frac{\pi}{4}}{2}.\frac{1 - 2\cos\frac{3\pi}{4}}{2}

    Substituting values from 105 we get desired result.

  22. We have to find the value of sin⁡23π24\sin \frac{23\pi}{24}

    sin⁡(π−π24)=sin⁡15∘2\sin \left(\pi - \frac{\pi}{24}\right) = \sin\frac{15^\circ}{2}

    sin⁡2A=12(1−cos⁡2A)=12(1−cos⁡15∘)\sin^2A = \frac{1}{2}(1 - \cos 2A) = \frac{1}{2}(1 - \cos15^\circ)

    =12(1−3+122)= \frac{1}{2}\left(1 - \frac{\sqrt{3} + 1}{2\sqrt{2}}\right)

    ∴sin⁡A=148−26−22\therefore \sin A = \frac{1}{4}\sqrt{8 - 2\sqrt{6} - 2\sqrt{2}}

  23. Given A=112∘30′∴2A=225∘A = 112^\circ30'\therefore 2A = 225^\circ

    cos⁡2A=cos⁡(180∘+45∘)=−12\cos 2A = \cos(180^\circ + 45^\circ) = -\frac{1}{\sqrt{2}}

    ∣sin⁡A∣=1−(−12)2=2+22|\sin A| = \sqrt{\frac{1 - \left(-\frac{1}{\sqrt{2}}\right)}{2}} = \sqrt{\frac{2 + \sqrt{2}}{2}}

    ∵\because A lies in 2nd quadrant ∴sin⁡A\therefore \sin A will be positive and cos⁡A\cos A will be negative.

    ∣cos⁡A∣=−2−22|\cos A| = -\frac{\sqrt{2 - \sqrt{2}}}{2}

  24. We have to prove that sin⁡224∘−sin⁡26∘=18(5−1)\sin^224^\circ - \sin^26^\circ = \frac{1}{8}(\sqrt{5} - 1)

    L.H.S. =sin⁡(24∘+6∘)sin⁡(24∘−6∘)=12.5−14= \sin(24^\circ + 6^\circ)\sin(24^\circ - 6^\circ) = \frac{1}{2}.\frac{\sqrt{5} - 1}{4}

    =18(5−1)== \frac{1}{8}(\sqrt{5} - 1) = R.H.S.

  25. We have to prove that tan⁡6∘.tan⁡42∘.tan⁡66∘.tan⁡78∘=1\tan6^\circ.\tan42^\circ.\tan66^\circ.\tan78^\circ = 1

    L.H.S. =sin⁡66∘6∘cos⁡66∘cos⁡6∘.sin⁡78∘sin⁡42∘cos⁡78∘cos⁡42∘= \frac{\sin66^\circ6^\circ}{\cos66^\circ\cos6^\circ}.\frac{\sin78^\circ\sin42^\circ}{\cos78^\circ\cos42^\circ}

    =cos⁡60∘−cos⁡72∘cos⁡60∘+cos⁡72∘.cos⁡36∘−cos⁡120∘cos⁡36∘+cos⁡120∘= \frac{\cos60^\circ - \cos72^\circ}{\cos60^\circ + \cos72^\circ}.\frac{\cos36^\circ - \cos120^\circ}{\cos36^\circ + \cos120^\circ}

    =1−2sin⁡18∘1+2sin⁡18∘.2cos⁡36∘+12cos⁡36∘−1= \frac{1 - 2\sin18^\circ}{1 + 2\sin18^\circ}.\frac{2\cos36^\circ + 1}{2\cos36^\circ - 1}

    =1−2(5−14)1+2(5−14).2.(5+14)+12.(5+14)−1= \frac{1 - 2\left(\frac{\sqrt{5} - 1}{4}\right)}{1 + 2\left(\frac{\sqrt{5} - 1}{4}\right)}.\frac{2.\left(\frac{\sqrt{5} + 1}{4}\right) + 1}{2.\left(\frac{\sqrt{5} + 1}{4}\right) - 1}

    =1== 1 = R.H.S.

  26. We have to prove that sin⁡47∘+sin⁡61∘−sin⁡11∘−sin⁡25∘=cos⁡7∘\sin47^\circ + \sin61^\circ - \sin 11^\circ - \sin25^\circ = \cos 7^\circ

    L.H.S. =2sin⁡54∘cos⁡7∘−2sin⁡18∘cos⁡7∘= 2\sin54^\circ\cos7^\circ - 2\sin18^\circ\cos7^\circ

    =2cos⁡7∘.2cos⁡36∘.sin⁡18∘=2cos⁡7∘.25+14.5−14= 2\cos7^\circ.2\cos36^\circ.\sin18^\circ = 2\cos7^\circ.2\frac{\sqrt{5} + 1}{4}.\frac{\sqrt{5} - 1}{4}

    =cos⁡7∘= \cos7^\circ

  27. We have to prove that sin⁡12∘sin⁡48∘sin⁡54∘=18\sin 12^\circ\sin48^\circ\sin54^\circ = \frac{1}{8}

    L.H.S. =12.2sin⁡48∘sin⁡12∘.sin⁡54∘= \frac{1}{2}.2\sin48^\circ\sin12^\circ.\sin54^\circ

    =12(cos⁡36∘−cos⁡60∘).cos⁡36∘= \frac{1}{2}(\cos 36^\circ - \cos60^\circ).\cos36^\circ

    =12(5+14−12).5+14= \frac{1}{2}\left(\frac{\sqrt{5} + 1}{4} - \frac{1}{2}\right).\frac{\sqrt{5} + 1}{4}

    =18== \frac{1}{8} = R.H.S.

  28. We have to prove that cot⁡14212∘=2+3−2−6\cot 142\frac{1}{2}^\circ = \sqrt{2} + \sqrt{3} - 2 - \sqrt{6}

    L.H.S. cos⁡14212∘=cot⁡(180∘−3712∘)=−cot⁡3712∘\cos 142\frac{1}{2}^\circ = \cot\left(180^\circ - 37\frac{1}{2}^\circ\right) = -\cot37\frac{1}{2}^\circ

    We know that tan⁡15∘=cot⁡75∘=3−13+1\tan 15^\circ = \cot 75^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}

    ∴−cot⁡3712∘=2+3−2−6=\therefore -\cot37\frac{1}{2}^\circ = \sqrt{2} + \sqrt{3} - 2 - \sqrt{6} = R.H.S.

  29. We have to prove that sin⁡248∘−cos⁡212∘=−5+18\sin^248^\circ - \cos^212^\circ = -\frac{\sqrt{5} + 1}{8}

    L.H.S. =12(2sin⁡248∘−2cos⁡212∘)= \frac{1}{2}\left(2\sin^248^\circ - 2\cos^212^\circ\right)

    =12(1−cos⁡96∘−1−cos⁡24∘)= \frac{1}{2}\left(1 - \cos96^\circ - 1 - \cos24^\circ\right)

    =−12(2cos⁡60∘cos⁡36∘)= -\frac{1}{2}\left(2\cos60^\circ\cos36^\circ\right)

    =−12.5+14=−5+18== -\frac{1}{2}.\frac{\sqrt{5} + 1}{4} = -\frac{\sqrt{5} + 1}{8} = R.H.S.

  30. We have to prove that 4(sin⁡24∘+cos⁡6∘)=3+154(\sin 24^\circ + \cos6^\circ) = \sqrt{3} + \sqrt{15}

    L.H.S. =4(sin⁡24∘+sin⁡84∘)=8sin⁡54∘cos⁡30∘=43sin⁡54∘= 4(\sin24^\circ + \sin84^\circ) = 8\sin54^\circ\cos30^\circ = 4\sqrt{3}\sin54^\circ

    =43(3sin⁡18∘−4sin⁡318∘)= 4\sqrt{3}(3\sin18^\circ - 4\sin^318^\circ)

    We know that sin⁡18∘=5−14\sin 18^\circ = \frac{\sqrt{5} - 1}{4}

    ∴43(3sin⁡18∘−4sin⁡318∘)=3+15=\therefore 4\sqrt{3}(3\sin18^\circ - 4\sin^318^\circ) = \sqrt{3} + \sqrt{15} = R.H.S.

  31. We have to prove that cot⁡6∘cot⁡42∘cot⁡66∘cot⁡78∘=1\cot6^\circ\cot42^\circ\cot66^\circ\cot78^\circ = 1

    L.H.S. =1tan⁡6∘tan⁡42∘tan⁡66∘tan⁡78∘= \frac{1}{\tan6^\circ\tan42^\circ\tan66^\circ\tan78^\circ}

    We know that tan⁡(60∘−x)tan⁡xtan⁡(60∘+x)=tan⁡3x\tan(60^\circ - x)\tan x\tan(60^\circ + x) = \tan 3x

    Putting x=18∘,x=18^\circ, we get

    tan⁡42∘tan⁡18∘tan⁡78∘=tan⁡54∘\tan42^\circ\tan18^\circ\tan78^\circ = \tan54^\circ

    Putting x=6∘,x=6^\circ, we get

    tan⁡54∘tan⁡6∘tan⁡66∘=tan⁡18∘\tan54^\circ\tan6^\circ\tan66^\circ = \tan18^\circ

    From these two, we derive that

    tan⁡6∘tan⁡42∘tan⁡66∘tan⁡78∘=1\tan6^\circ\tan42^\circ\tan66^\circ\tan78^\circ = 1

  32. We have to prove that tan⁡12∘tan⁡24∘tan⁡48∘tan⁡84∘=1\tan12^\circ\tan24^\circ\tan48^\circ\tan84^\circ = 1

    We know that tan⁡(60∘−x)tan⁡xtan⁡(60∘+x)=tan⁡3x\tan(60^\circ - x)\tan x\tan(60^\circ + x) = \tan 3x

    Putting x=12∘,x= 12^\circ, we get

    tan⁡48∘tan⁡12∘tan⁡72∘=tan⁡36∘\tan48^\circ\tan 12^\circ\tan72^\circ = \tan 36^\circ

    Putthing x=24∘,x = 24^\circ, we get

    tan⁡36∘tan⁡24∘tan⁡84∘=tan⁡72∘\tan36^\circ\tan24^\circ\tan84^\circ = \tan72^\circ

    From these two, we derive that

    tan⁡12∘tan⁡24∘tan⁡48∘tan⁡84∘=1\tan12^\circ\tan24^\circ\tan48^\circ\tan84^\circ = 1

  33. We have to prove that sin⁡6∘sin⁡42∘sin⁡66∘sin⁡78∘=116\sin6^\circ\sin42^\circ\sin66^\circ\sin78^\circ = \frac{1}{16}

    L.H.S. =sin⁡6∘sin⁡66∘sin⁡42∘sin⁡78∘= \sin6^\circ\sin66^\circ\sin42^\circ\sin78^\circ

    =14(cos⁡60∘−cos⁡72∘)(cos⁡36∘−cos⁡120∘)= \frac{1}{4}(\cos60^\circ - \cos72^\circ)(\cos 36^\circ - \cos120^\circ)

    =14(12−cos⁡72∘)(cos⁡36∘+12)= \frac{1}{4}\left(\frac{1}{2} - \cos72^\circ\right)\left(\cos36^\circ + \frac{1}{2}\right)

    =116(1−2cos⁡72∘)(2cos⁡36∘+1)= \frac{1}{16}(1 - 2\cos72^\circ)(2\cos36^\circ + 1)

    =116[1+2cos⁡36∘−2cos⁡72∘−4cos⁡36∘cos⁡72∘]= \frac{1}{16}[1 + 2\cos36^\circ - 2\cos72^\circ - 4\cos36^\circ\cos72^\circ]

    =116+18[cos⁡36∘−cos⁡72∘−cos⁡108∘−cos⁡36∘]= \frac{1}{16} + \frac{1}{8}[\cos36^\circ - \cos72^\circ - \cos 108^\circ - \cos36^\circ]

    =116+18[cos⁡72∘+∘108∘]= \frac{1}{16} + \frac{1}{8}[\cos72^\circ + \circ108^\circ]

    =116+18[cos⁡72∘+cos⁡(180∘−72∘)]= \frac{1}{16} + \frac{1}{8}[\cos72^\circ + \cos(180^\circ - 72^\circ)]

    =116= \frac{1}{16}

  34. We have to prove that sin⁡π5sin⁡2π5sin⁡3π5sin⁡4π5=516\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}

    L.H.S. =sin⁡π5sin⁡2π5sin⁡(π−2π5)sin⁡(π−π5)= \sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\left(\pi - \frac{2\pi}{5}\right)\sin\left(\pi - \frac{\pi}{5}\right)

    =sin⁡2π5sin⁡22π5=sin⁡218∘sin⁡236∘=(5−14)2(1410−25)2= \sin^2\frac{\pi}{5}\sin^2\frac{2\pi}{5} = \sin^218^\circ\sin^236^\circ = \left(\frac{\sqrt{5} - 1}{4}\right)^2\left(\frac{1}{4}\sqrt{10 - 2\sqrt{5}}\right)^2

    =516== \frac{5}{16} = R.H.S.

  35. We have to prove that cos⁡36∘cos⁡72∘cos⁡108∘cos⁡144∘=116\cos36^\circ\cos72^\circ\cos108^\circ\cos144^\circ = \frac{1}{16}

    L.H.S. =cos⁡36∘cos⁡72∘cos⁡(180∘−72∘)cos⁡(180∘−36∘)= \cos36^\circ\cos72^\circ\cos(180^\circ - 72^\circ)\cos(180^\circ - 36^\circ)

    =cos⁡236∘cos⁡272∘= \cos^236^\circ\cos^272^\circ

    cos⁡36∘=5+14,cos⁡72∘=2cos⁡236∘−1\cos 36^\circ = \frac{\sqrt{5} + 1}{4}, \cos72^\circ = 2\cos^236^\circ - 1

    Thus, cos⁡236∘cos⁡272∘=116\cos^236^\circ\cos^272^\circ = \frac{1}{16}

  36. We have to prove that cos⁡π15cos⁡2π15cos⁡3π15cos⁡4π15cos⁡5π15cos⁡6π15cos⁡7π15=127\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15} = \frac{1}{2^7}

    L.H.S. =12sin⁡π152sin⁡π15cos⁡π15cos⁡2π15cos⁡3π15cos⁡4π15cos⁡5π15cos⁡6π15cos⁡7π15= \frac{1}{2\sin\frac{\pi}{15}}2\sin\frac{\pi}{15}\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}

    =12sin⁡π15sin⁡2π15cos⁡2π15cos⁡3π15cos⁡4π15cos⁡5π15cos⁡6π15cos⁡7π15= \frac{1}{2\sin\frac{\pi}{15}}\sin\frac{2\pi}{15}\cos\frac{2\pi}{15}\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}

    =122sin⁡π15sin⁡4π15cos⁡3π15cos⁡4π15cos⁡5π15cos⁡6π15cos⁡7π15= \frac{1}{2^2\sin\frac{\pi}{15}}\sin\frac{4\pi}{15}\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}

    =123sin⁡π15sin⁡8π15cos⁡3π15cos⁡5π15cos⁡6π15cos⁡7π15= \frac{1}{2^3\sin\frac{\pi}{15}}\sin\frac{8\pi}{15}\cos\frac{3\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}

    Now, sin⁡8π15=sin⁡(π−7π15)=sin⁡7π15,\sin\frac{8\pi}{15} = \sin\left(\pi - \frac{7\pi}{15}\right) = \sin\frac{7\pi}{15}, therefore

    =124sin⁡π152sin⁡7π15cos⁡3π15cos⁡5π15cos⁡6π15cos⁡7π15= \frac{1}{2^4\sin\frac{\pi}{15}}2\sin\frac{7\pi}{15}\cos\frac{3\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}

    =124sin⁡π15sin⁡14π15cos⁡3π15cos⁡5π15cos⁡6π15= \frac{1}{2^4\sin\frac{\pi}{15}}\sin\frac{14\pi}{15}\cos\frac{3\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}

    Now sin⁡14π15=sin⁡(π−π15)=sin⁡π15,\sin\frac{14\pi}{15} = \sin\left(\pi - \frac{\pi}{15}\right) = \sin\frac{\pi}{15}, therefore

    =124cos⁡3π15cos⁡5π15cos⁡6π15= \frac{1}{2^4}\cos\frac{3\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}

    =125sin⁡3π152sin⁡3π15cos⁡3π15cos⁡5π15cos⁡6π15= \frac{1}{2^5\sin\frac{3\pi}{15}}2\sin\frac{3\pi}{15}\cos\frac{3\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}

    =126sin⁡3π152sin⁡6π15cos⁡6π15cos⁡π3= \frac{1}{2^6\sin\frac{3\pi}{15}}2\sin\frac{6\pi}{15}\cos\frac{6\pi}{15}\cos\frac{\pi}{3}

    =126sin⁡3π15sin⁡12π15cos⁡π3= \frac{1}{2^6\sin\frac{3\pi}{15}}\sin\frac{12\pi}{15}\cos\frac{\pi}{3}

    Similarly sin⁡12π15=sin⁡3π15\sin\frac{12\pi}{15} = \sin \frac{3\pi}{15}

    =126cos⁡π3=127== \frac{1}{2^6}\cos\frac{\pi}{3} = \frac{1}{2^7} = R.H.S.

  37. We have to prove that cos⁡π65cos⁡2π65cos⁡4π65cos⁡8π65cos⁡16π65cos⁡32π65=164\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65} = \frac{1}{64}

    =12sin⁡π652sin⁡π65cos⁡π65cos⁡2π65cos⁡4π65cos⁡8π65cos⁡16π65cos⁡32π65= \frac{1}{2\sin\frac{\pi}{65}}2\sin\frac{\pi}{65}\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}

    =12sin⁡π65sin⁡2π65cos⁡2π65cos⁡4π65cos⁡8π65cos⁡16π65cos⁡32π65= \frac{1}{2\sin\frac{\pi}{65}}\sin\frac{2\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}

    =122sin⁡π652sin⁡2π65cos⁡2π65cos⁡4π65cos⁡8π65cos⁡16π65cos⁡32π65= \frac{1}{2^2\sin\frac{\pi}{65}}2\sin\frac{2\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}

    =122sin⁡π65sin⁡4π65cos⁡4π65cos⁡8π65cos⁡16π65cos⁡32π65= \frac{1}{2^2\sin\frac{\pi}{65}}\sin\frac{4\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}

    =123sin⁡π652sin⁡4π65cos⁡4π65cos⁡8π65cos⁡16π65cos⁡32π65= \frac{1}{2^3\sin\frac{\pi}{65}}2\sin\frac{4\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}

    Proceeding similalry we find that above is equal to

    127sin⁡π65sin⁡64π65\frac{1}{2^7\sin\frac{\pi}{65}}\sin\frac{64\pi}{65}

    However, sin⁡64π65=sin⁡π65,\sin\frac{64\pi}{65} = \sin\frac{\pi}{65}, therefore

    cos⁡π65cos⁡2π65cos⁡4π65cos⁡8π65cos⁡16π65cos⁡32π65=164\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65} = \frac{1}{64}

  38. Given, tan⁡A2=a−ba+btan⁡B2\tan \frac{A}{2} = \sqrt{\frac{a - b}{a + b}}\tan \frac{B}{2}

    Now, cos⁡A=1−tan⁡2A21+tan⁡2A2=1−a−ba+btan⁡2B21+a−ba+btan⁡2B2\cos A = \frac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}} = \frac{1 - \frac{a - b}{a + b}\tan^2\frac{B}{2}}{1 + \frac{a - b}{a + b}\tan^2\frac{B}{2}}

    =(a+b)cos⁡2B2−(a−b)sin⁡2B2(a+b)cos⁡2B2+(a−b)sin⁡2B2= \frac{(a + b)\cos^2\frac{B}{2} - (a - b)\sin^2\frac{B}{2}}{(a + b)\cos^2\frac{B}{2} + (a - b)\sin^2\frac{B}{2}}

    =acos⁡B+ba+bcos⁡B= \frac{a\cos B + b}{a + b\cos B}

  39. This problem is similar to previous problem with a=1,e=ba = 1, e = b and has been left as an exercise.

  40. Given sin⁡A+sin⁡B=a\sin A + \sin B = a and cos⁡A+cos⁡B=b,\cos A + \cos B = b, we have to prove that sin⁡(A+B)=2aba2+b2\sin(A + B) = \frac{2ab}{a^2 + b^2}

    2ab=2(sin⁡A+sin⁡B)(cos⁡A+cos⁡B)=2sin⁡Acos⁡A+2sin⁡Acos⁡B+2sin⁡Bcos⁡A+2sin⁡Bcos⁡B=sin⁡2A+sin⁡2B+2sin⁡(A+B)2ab = 2(\sin A + \sin B)(\cos A + \cos B) = 2\sin A\cos A + 2\sin A\cos B + 2\sin B\cos A + 2\sin B\cos B = \sin 2A + \sin 2B + 2\sin(A + B)

    =2sin⁡(A+B)[cos⁡(B−A)+1]= 2\sin(A + B)[\cos(B - A) + 1]

    a2+b2=sin⁡2A+sin⁡2B+2sin⁡Asin⁡B+cos⁡2A+cos⁡2B+2cos⁡Acos⁡Ba^2 + b^2 = \sin^2A + \sin^2B + 2\sin A\sin B + \cos^2A + \cos^2B + 2\cos A\cos B

    =2+2cos⁡(B−A)= 2 + 2\cos(B - A)

    ∴sin⁡(A+B)=2aba2+b2\therefore \sin(A + B) = \frac{2ab}{a^2 + b^2}

  41. Given sin⁡A+sin⁡B=a\sin A + \sin B = a and cos⁡A+cos⁡B=b,\cos A + \cos B = b, we have to prove that cos⁡(A−B)=12(a2+b2−2)\cos(A - B) = \frac{1}{2}(a^2 + b^2 - 2)

    From previous problem, 2cos⁡(B−A)=a2+b2−2⇒cos⁡(A−B)=12(a2+b2−2)2\cos(B - A) = a^2 + b^2 - 2 \Rightarrow \cos(A - B) = \frac{1}{2}(a^2 + b^2 - 2)

  42. Let us solve these one by one.

    1. Given AA and BB be two different roots of equation acos⁡θ+bsin⁡θ=ca\cos\theta + b\sin\theta = c

      acos⁡A+bsin⁡A=ca\cos A + b\sin A = c and acos⁡B+bsin⁡B=ca\cos B + b\sin B = c

      ⇒a(cos⁡A−cos⁡B)+b(sin⁡A−sin⁡B)=0\Rightarrow a(\cos A - \cos B) + b(\sin A - \sin B) = 0

      b(sin⁡A−sin⁡B)=a(cos⁡A−cos⁡B)b(\sin A - \sin B) = a(\cos A - \cos B)

      b.2cos⁡A+B2sin⁡A−B2=a.2sin⁡A+B2sin⁡A−B2b.2\cos\frac{A + B}{2}\sin\frac{A - B}{2} = a.2\sin\frac{A + B}{2}\sin \frac{A - B}{2}

      ⇒tan⁡A+B2=ba\Rightarrow \tan\frac{A + B}{2} = \frac{b}{a}

      tan⁡A+B=2tan⁡A+B21−tan⁡2A+B2=2aba2+b2\tan{A + B} = \frac{2\tan\frac{A + B}{2}}{1 - \tan^2\frac{A + B}{2}} = \frac{2ab}{a^2 + b^2}

    2. We have tan⁡(A+B)=2aba2+b2\tan(A + B) = \frac{2ab}{a^2 + b^2}

      ∴cos⁡(A+B)=a2−b2a2+b2\therefore \cos(A + B) = \frac{a^2 - b^2}{a^2 + b^2}

  43. Given cos⁡A+cos⁡B=13\cos A + \cos B = \frac{1}{3} and sin⁡A+sin⁡B=14,\sin A + \sin B = \frac{1}{4}, we have to prove that cos⁡A−B2=±524\cos \frac{A - B}{2} = \pm\frac{5}{24}

    Squaring and adding (cos⁡2A+sin⁡2A)+(cos⁡2B+sin⁡B)+2(cos⁡Acos⁡B+sin⁡Asin⁡B)=19+116(\cos^2A + \sin^2A) + (\cos^2B + \sin^B) + 2(\cos A\cos B + \sin A\sin B) = \frac{1}{9} + \frac{1}{16}

    2+2cos⁡(A−B)=251442 + 2\cos(A - B) = \frac{25}{144}

    4cos⁡2A−B2=25144⇒cos⁡A−B2=±5244\cos^2\frac{A - B}{2} = \frac{25}{144} \Rightarrow \cos\frac{A - B}{2} = \pm\frac{5}{24}

  44. Given 2tan⁡A2=tan⁡B2,2\tan \frac{A}{2} = \tan \frac{B}{2}, we have to prove that cos⁡A=3+5cos⁡B5+3cos⁡B\cos A = \frac{3 + 5\cos B}{5 + 3\cos B}

    tan⁡A2=12tan⁡B2\tan\frac{A}{2} = \frac{1}{2}\tan\frac{B}{2}

    cos⁡A=1−tan⁡2A21+tan⁡2A2=1−tan⁡2B241+tan⁡2B24\cos A = \frac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}} = \frac{1 - \frac{\tan^2\frac{B}{2}}{4}}{1 + \frac{\tan^2\frac{B}{2}}{4}}

    =4−tan⁡2B24+tan⁡2B2=3+5.1−tan⁡2B21+tan⁡2B25+31−tan⁡2B21+tan⁡2B2= \frac{4 - \tan^2\frac{B}{2}}{4 + \tan^2\frac{B}{2}} = \frac{3 + 5.\frac{1 - \tan^2\frac{B}{2}}{1 + \tan^2\frac{B}{2}}}{5 + 3\frac{1 - \tan^2\frac{B}{2}}{1 + \tan^2\frac{B}{2}}}

    =3+5cos⁡B5+3cos⁡B== \frac{3 + 5\cos B}{5 + 3\cos B} = R.H.S.

  45. Given sin⁡A=45\sin A = \frac{4}{5} and cos⁡B=513,\cos B = \frac{5}{13}, we have to prove that one value of cos⁡A−B2=865\cos \frac{A - B}{2} = \frac{8}{\sqrt{65}}

    cos⁡A=35\cos A = \frac{3}{5} and sin⁡B=1213\sin B = \frac{12}{13}

    cos⁡2A−B2=1+cos⁡(A−B)2\cos^2\frac{A - B}{2} = \frac{1 + \cos(A - B)}{2}

    cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B=1565+4865=6365\cos(A - B) = \cos A\cos B + \sin A\sin B = \frac{15}{65} + \frac{48}{65} = \frac{63}{65}

    1+cos⁡(A−B)2=1282.65\frac{1 + \cos(A - B)}{2} = \frac{128}{2.65}

    cos⁡A−B2=±865\cos\frac{A - B}{2} = \pm\frac{8}{\sqrt{65}}

  46. Given, sec⁡(A+B)+sec⁡(A−B)=2sec⁡A,\sec(A + B) + \sec(A - B) = 2\sec A, we have to prove that cos⁡B=±2cos⁡B2\cos B = \pm \sqrt{2}\cos \frac{B}{2}

    L.H.S. =1cos⁡(A+B)+1cos⁡(A−B)=cos⁡(A−B)+cos⁡(A+B)cos⁡(A−B)cos⁡(A+B)= \frac{1}{\cos(A + B)} + \frac{1}{\cos(A - B)} = \frac{\cos(A - B) + \cos (A + B)}{\cos(A - B)\cos(A + B)}

    =4(cos⁡Acos⁡B)cos⁡2A+cos⁡2B= \frac{4(\cos A\cos B)}{\cos 2A + \cos 2B}

    2cos⁡Acos⁡Bcos⁡2A+cos⁡2B=1cos⁡A\frac{2\cos A\cos B}{\cos 2A + \cos 2B} = \frac{1}{\cos A}

    2cos⁡2Acos⁡B=2cos⁡2A−1+2cos⁡2B−12\cos^2A\cos B = 2\cos^2A - 1 + 2\cos^2B - 1

    2cos⁡2A(cos⁡B−1)=2(cos⁡2B−1)2\cos^2A(\cos B - 1) = 2(\cos^2B - 1)

    cos⁡2A=cos⁡B+1=2cos⁡2B2\cos^2A = \cos B + 1 = 2\cos^2\frac{B}{2}

    cos⁡A=±2cos⁡B2\cos A = \pm\sqrt{2}\cos\frac{B}{2}

  47. Given cos⁡θ=cos⁡αcos⁡β1−sin⁡αsin⁡β,\cos \theta = \frac{\cos\alpha\cos\beta}{1 - \sin\alpha\sin\beta}, we have to prove that one of the values of tan⁡θ2\tan \frac{\theta}{2} is tan⁡α2−tan⁡β21−tan⁡α2tan⁡β2\frac{\tan \frac{\alpha}{2} - \tan\frac{\beta}{2}}{1 - \tan\frac{\alpha}{2}\tan\frac{\beta}{2}}

    tan⁡2θ2=1−cos⁡θ1+cos⁡θ\tan^2\frac{\theta}{2} = \frac{1 - \cos\theta}{1 + \cos\theta}

    =1−cos⁡αcos⁡β1−sin⁡αsin⁡β1+cos⁡αcos⁡β1−sin⁡αsin⁡β= \frac{1 - \frac{\cos\alpha\cos\beta}{1 - \sin\alpha\sin\beta}}{1 + \frac{\cos\alpha\cos\beta}{1 - \sin\alpha\sin\beta}}

    =1−(cos⁡αcos⁡β+sin⁡αsin⁡β)1+(cos⁡αcos⁡β−sin⁡αsin⁡β)= \frac{1 - (\cos\alpha\cos\beta + \sin\alpha\sin\beta)}{1 + (\cos\alpha\cos\beta - \sin\alpha\sin\beta)}

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

    =2sin⁡2α−β22cos⁡2α+β2= \frac{2\sin^2\frac{\alpha - \beta}{2}}{2\cos^2\frac{\alpha + \beta}{2}}

    tan⁡θ2=sin⁡α−β2cos⁡α+β2\tan\frac{\theta}{2} = \frac{\sin\frac{\alpha - \beta}{2}}{\cos\frac{\alpha + \beta}{2}}

    =sin⁡α2cos⁡β2−cos⁡α2cos⁡β2cos⁡α2cos⁡β2−sin⁡α2sin⁡β2= \frac{\sin\frac{\alpha}{2}\cos\frac{\beta}{2} - \cos\frac{\alpha}{2}\cos\frac{\beta}{2}}{\cos\frac{\alpha}{2}\cos\frac{\beta}{2} - \sin\frac{\alpha}{2}\sin\frac{\beta}{2}}

    Dividing both numerator and denominator by cos⁡α2cos⁡β2\cos\frac{\alpha}{2}\cos\frac{\beta}{2}

    tan⁡θ2=tan⁡α2−tan⁡β21−tan⁡α2tan⁡β2\tan\frac{\theta}{2} = \frac{\tan \frac{\alpha}{2} - \tan\frac{\beta}{2}}{1 - \tan\frac{\alpha}{2}\tan\frac{\beta}{2}}

  48. Given tan⁡α=sin⁡θsin⁡ϕcos⁡θ+cos⁡ϕ,\tan\alpha = \frac{\sin\theta\sin\phi}{\cos\theta + \cos\phi}, we have to prove that one of the values of tan⁡α2\tan\frac{\alpha}{2} is tan⁡θ2tan⁡ϕ2\tan\frac{\theta}{2}\tan\frac{\phi}{2}

    ⇒2tan⁡α21−tan⁡2α2=2tan⁡θ21+tan⁡2θ22tan⁡ϕ21+tan⁡2ϕ21−tan⁡2θ21+tan⁡2θ2+1−tan⁡2ϕ21+tan⁡2ϕ2\Rightarrow \frac{2\tan\frac{\alpha}{2}}{1 - \tan^2\frac{\alpha}{2}} = \frac{\frac{2\tan\frac{\theta}{2}}{1 + \tan^2\frac{\theta}{2}}\frac{2\tan\frac{\phi}{2}}{1 + \tan^2\frac{\phi}{2}}}{\frac{1 - \tan^2\frac{\theta}{2}}{1 + \tan^2\frac{\theta}{2}} + \frac{1 - \tan^2\frac{\phi}{2}}{1 + \tan^2\frac{\phi}{2}}}

    =4tan⁡θ2tan⁡ϕ21+tan⁡2ϕ2−tan⁡2θ2−tan⁡2ϕ2.tan⁡2θ2+1+tan⁡2θ2−tan⁡2ϕ2−tan⁡2ϕ2.tan⁡2θ2= \frac{4\tan\frac{\theta}{2}\tan\frac{\phi}{2}}{1 + \tan^2\frac{\phi}{2} - \tan^2\frac{\theta}{2} - \tan^2\frac{\phi}{2}.\tan^2\frac{\theta}{2} + 1 + \tan^2\frac{\theta}{2} -\tan^2\frac{\phi}{2} - \tan^2\frac{\phi}{2}.\tan^2\frac{\theta}{2}}

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

    Solving this quadratic equationin tan⁡α2\tan\frac{\alpha}{2} we obtain the desired result.

  49. Given cos⁡θ=cos⁡α+cos⁡β1+cos⁡αcos⁡β,\cos\theta = \frac{\cos\alpha + \cos\beta}{1 + \cos\alpha\cos\beta}, we have to prove that one of the values of tan⁡θ2\tan\frac{\theta}{2} is tan⁡α2tan⁡β2\tan\frac{\alpha}{2}\tan\frac{\beta}{2}

    cos⁡θ=cos⁡α+cos⁡β1+cos⁡αcos⁡β\cos\theta = \frac{\cos\alpha + \cos\beta}{1 + \cos\alpha\cos\beta}

    1−tan⁡2θ21+tan⁡2θ2=cos⁡α+cos⁡β1+cos⁡αcos⁡β\frac{1 - \tan^2\frac{\theta}{2}}{1 + \tan^2\frac{\theta}{2}} = \frac{\cos\alpha + \cos\beta}{1 + \cos\alpha\cos\beta}

    tan⁡2θ2=1−cos⁡αcos⁡β−cos⁡α+cos⁡β1−cos⁡αcos⁡β+cos⁡α−cos⁡β\tan^2\frac{\theta}{2} = \frac{1 - \cos\alpha\cos\beta - \cos\alpha + \cos\beta}{1 - \cos\alpha\cos\beta + \cos\alpha - \cos\beta}

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

    tan⁡2θ2=tan⁡2α2cot⁡2β2\tan^2\frac{\theta}{2} = \tan^2\frac{\alpha}{2}\cot^2\frac{\beta}{2}

    tan⁡θ2=±tan⁡α2cot⁡β2\tan\frac{\theta}{2} = \pm\tan\frac{\alpha}{2}\cot\frac{\beta}{2}