11. Transformation Formulae Solutions#

  1. Given, sin⁡75∘−sin⁡15∘cos⁡75∘+cos⁡15∘\frac{\sin 75^\circ - \sin 15^\circ}{\cos 75^\circ + \cos 15^\circ}

    =2cos⁡75∘+15∘2sin⁡75∘−15∘22cos⁡75∘+15∘2cos⁡75∘−15∘2= \frac{2\cos \frac{75^\circ + 15^\circ}{2}\sin \frac{75^\circ - 15^\circ}{2}}{2\cos \frac{75^\circ + 15^\circ}{2}\cos \frac{75^\circ - 15^\circ}{2}}

    =2cos⁡45∘sin⁡30∘2cos⁡45∘cos⁡30∘=tan⁡30∘=13= \frac{2\cos 45^\circ\sin30^\circ}{2\cos45^\circ\cos30^\circ} = \tan30^\circ = \frac{1}{\sqrt{3}}

  2. Given, (cos⁡θ−cos⁡2θ)(sin⁡8θ+sin⁡2θ)(sin⁡5θ−sin⁡θ)(cos⁡4θ−cos⁡6θ)=2sin⁡2θsin⁡θ.2sin⁡5θcos⁡3θ2cos⁡3θsin⁡2θ.2sin⁡5θsin⁡θ\frac{(\cos \theta - \cos 2\theta)(\sin 8\theta + \sin 2\theta)}{(\sin 5\theta - \sin\theta)(\cos 4\theta - \cos 6\theta)} = \frac{2\sin2\theta\sin\theta.2\sin5\theta\cos3\theta}{2\cos3\theta\sin2\theta.2\sin5\theta\sin\theta}

    =1= 1

  3. We have to prove, sin⁡7θ−sin⁡5θcos⁡7θ+cos⁡5θ=tan⁡θ\frac{\sin7\theta - \sin5\theta}{\cos7\theta + \cos5\theta} = \tan\theta

    L.H.S. =2cos⁡6θsin⁡θ2cos⁡6θcos⁡θ=tan⁡θ= \frac{2\cos6\theta\sin\theta}{2\cos6\theta\cos\theta} = \tan\theta

  4. We have to prove, cos⁡6θ−cos⁡4θsin⁡6θ+sin⁡4θ=−tan⁡θ\frac{\cos6\theta - \cos4\theta}{\sin6\theta + \sin4\theta} = -\tan\theta

    L.H.S. =2sin⁡5θsin⁡(−θ)2sin⁡5θcos⁡θ=−tan⁡θ= \frac{2\sin5\theta\sin(-\theta)}{2\sin5\theta\cos\theta} = -\tan\theta

  5. We have to prove, sin⁡A+sin⁡3Acos⁡A+cos⁡3A=tan⁡2A\frac{\sin A + \sin 3A}{\cos A + \cos 3A} = \tan 2A

    L.H.S. =2sin⁡2Acos⁡(−A)2cos⁡2Acos⁡(−A)=tan⁡2A= \frac{2\sin2A\cos(-A)}{2\cos2A\cos(-A)} = \tan 2A

  6. We have to prove, sin⁡7A−sin⁡Asin⁡8A−sin⁡2A=cos⁡4Asec⁡5A\frac{\sin 7A - \sin A}{\sin 8A - \sin 2A} = \cos 4A\sec 5A

    L.H.S. =2cos⁡4Asin⁡3A2cos⁡5Asin⁡3A=cos⁡4Asec⁡5A= \frac{2\cos4A\sin3A}{2\cos5A\sin3A} = \cos4A\sec5A

  7. We have to prove, cos⁡2B+cos⁡2Acos⁡2B−cos⁡2A=cot⁡(A+B)cot⁡(A−B)\frac{\cos 2B + \cos 2A}{\cos 2B - \cos 2A} = \cot(A + B)\cot(A - B)

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

  8. We have to prove, sin⁡2A+sin⁡2Bsin⁡2A−sin⁡2B=tan⁡(A+B)tan⁡(A−B)\frac{\sin 2A + \sin 2B}{\sin 2A - \sin 2B} = \frac{\tan(A + B)}{\tan(A - B)}

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

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

  9. We have to prove, sin⁡A+sin⁡2Acos⁡A−cos⁡2A=cot⁡A2\frac{\sin A + \sin 2A}{\cos A - \cos 2A} = \cot \frac{A}{2}

    L.H.S. =2sin⁡3A2cos⁡A22sin⁡3A2sin⁡A2= \frac{2\sin\frac{3A}{2}\cos\frac{A}{2}}{2\sin\frac{3A}{2}\sin\frac{A}{2}}

    =cot⁡A2= \cot\frac{A}{2}

  10. We have to prove, sin⁡5A−sin⁡3Acos⁡3A+cos⁡5A=tan⁡A\frac{\sin 5A - \sin 3A}{\cos 3A + \cos 5A} = \tan A

    L.H.S. =2cos⁡4Asin⁡A2cos⁡4Acos⁡A=tan⁡A= \frac{2\cos4A\sin A}{2\cos4A\cos A} = \tan A

  11. We have to prove, cos⁡2B−cos⁡2Asin⁡2B+sin⁡2A=tan⁡(A−B)\frac{\cos 2B - \cos 2A}{\sin 2B + \sin 2A} = \tan(A - B)

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

  12. We have to prove, cos⁡(A+B)+sin⁡(A−B)=2sin⁡(45∘+A)cos⁡(45∘+B)\cos (A + B) + \sin(A - B) = 2\sin(45^\circ + A)\cos(45^\circ + B)

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

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

    =2(12sin⁡A+12cos⁡A)(12cos⁡B−12sin⁡B)= 2(\frac{1}{\sqrt{2}}\sin A + \frac{1}{\sqrt{2}}\cos A)(\frac{1}{\sqrt{2}}\cos B - \frac{1}{\sqrt{2}}\sin B)

    =2(sin⁡Acos⁡45∘+sin⁡45∘sin⁡A)(cos⁡45∘cos⁡B−sin⁡45∘sin⁡B)= 2(\sin A\cos 45^\circ + \sin 45^\circ\sin A)(\cos 45^\circ\cos B - \sin 45^\circ\sin B)

    =2sin⁡(45∘+A)cos⁡(45∘+B)= 2\sin(45^\circ + A)\cos(45^\circ + B)

  13. We have to prove, cos⁡3A−cos⁡Asin⁡3A−sin⁡A+cos⁡2A−cos⁡4Asin⁡4A−sin⁡2A=sin⁡Acos⁡2Acos⁡3A\frac{\cos 3A - \cos A}{\sin 3A - \sin A} + \frac{\cos 2A - \cos 4A}{\sin 4A - \sin 2A} = \frac{\sin A}{\cos 2A\cos 3A}

    L.H.S. =−2sin⁡2Asin⁡A2cos⁡2Asin⁡A+2sin⁡3Asin⁡A2cos⁡3Asin⁡A= \frac{-2\sin 2A\sin A}{2\cos 2A\sin A} + \frac{2\sin 3A\sin A}{2\cos 3A\sin A}

    =−sin⁡2Acos⁡2A+sin⁡3Acos⁡3A= \frac{-\sin 2A}{\cos 2A} + \frac{\sin 3A}{\cos 3A}

    =cos⁡2Asin⁡3A−sin⁡2Acos⁡3Acos⁡2Acos⁡3A=sin⁡(3A−2A)cos⁡2Acos⁡3A= \frac{\cos 2A\sin 3A - \sin 2A\cos 3A}{\cos 2A\cos 3A} = \frac{\sin(3A - 2A)}{\cos 2A\cos 3A}

    =sin⁡Acos⁡3Acos⁡3A= \frac{\sin A}{\cos 3A\cos 3A}

  14. Given, sin⁡(4A−2B)+sin⁡(4B−2A)cos⁡(4A−2B)+cos⁡(4B−2A)=tan⁡(A+B)\frac{\sin (4A - 2B) + \sin (4B - 2A)}{\cos (4A - 2B) + \cos (4B - 2A)} = \tan(A + B) L.H.S. =2sin⁡(A+B)cos⁡3(A−B)2cos⁡(A+B)cos⁡3(A−B)=tan⁡(A+B)= \frac{2\sin(A + B)\cos3(A - B)}{2\cos(A + B)\cos3(A - B)} = \tan(A + B)

  15. We have to prove, tan⁡5θ+tan⁡3θtan⁡5θ−tan⁡3θ=4cos⁡2θcos⁡4θ\frac{\tan 5\theta + \tan 3\theta}{\tan 5\theta - \tan 3\theta} = 4\cos 2\theta\cos 4\theta

    L.H.S. =sin⁡5θcos⁡5θ+sin⁡4θcos⁡3θsin⁡5θcos⁡5θ−sin⁡4θcos⁡3θ= \frac{\frac{\sin5\theta}{\cos5\theta} + \frac{\sin4\theta}{\cos3\theta}}{\frac{\sin5\theta}{\cos5\theta} - \frac{\sin4\theta}{\cos3\theta}}

    =sin⁡5θcos⁡3θ+sin⁡3θcos⁡5θsin⁡5θcos⁡3θ−sin⁡3θcos⁡5θ= \frac{\sin5\theta\cos3\theta + \sin3\theta\cos5\theta}{\sin5\theta\cos3\theta - \sin3\theta\cos5\theta}

    =sin⁡8θsin⁡2θ=2sin⁡4θcos⁡θsin⁡2θ= \frac{\sin8\theta}{\sin2\theta} = \frac{2\sin4\theta\cos\theta}{\sin2\theta}

    =4sin⁡2θcos⁡2θcos⁡4θsin⁡2θ=4cos⁡2θcos⁡4θ= \frac{4\sin2\theta\cos2\theta\cos4\theta}{\sin2\theta} = 4\cos2\theta\cos4\theta

  16. We have to prove, cos⁡3θ+2cos⁡5θ+cos⁡7θcos⁡θ+2cos⁡3θ+cos⁡5θ=cos⁡2θ−sin⁡2θtan⁡3θ\frac{\cos 3\theta + 2\cos5\theta + \cos 7\theta}{\cos\theta + 2\cos3\theta + \cos 5\theta} = \cos 2\theta - \sin 2\theta\tan 3\theta

    Adding first and last terms of numerator and denominator, we have

    L.H.S. =2cos⁡5θcos⁡2θ+2cos⁡5θ2cos⁡3θcos⁡2θ+2cos⁡3θ= \frac{2\cos5\theta\cos2\theta + 2\cos5\theta}{2\cos3\theta\cos2\theta + 2\cos3\theta}

    =cos⁡5θ(cos⁡2θ+1)cos⁡3θ(cos⁡2θ+1)= \frac{\cos5\theta(\cos2\theta + 1)}{\cos3\theta(\cos2\theta + 1)}

    =cos⁡3θcos⁡2θ−sin⁡3θsin⁡2θcos⁡3θ= \frac{\cos3\theta\cos2\theta - \sin3\theta\sin2\theta}{\cos3\theta}

    =cos⁡2θ−sin⁡2θtan⁡3θ= \cos2\theta - \sin2\theta\tan3\theta

  17. We have to prove, sin⁡A+sin⁡3A+sin⁡5A+sin⁡7Acos⁡A+cos⁡3A+cos⁡5A+cos⁡7A=tan⁡4A\frac{\sin A + \sin 3A + \sin 5A + \sin 7A}{\cos A + \cos 3A + \cos 5A + \cos 7A} = \tan 4A

    Pairing first and fourth term and second and third term in numerator and denominator, we get

    L.H.S. =2sin⁡4Acos⁡3A+2sin⁡4Acos⁡A2cos⁡4Acos⁡3A+2cos⁡4Acos⁡A= \frac{2\sin4A\cos3A + 2\sin4A\cos A}{2\cos4A\cos3A + 2\cos4A\cos A}

    =2sin⁡4A(cos⁡3A+cos⁡A)2cos⁡4A(cos⁡3A+cos⁡A)= \frac{2\sin4A(\cos 3A + \cos A)}{2\cos4A(\cos 3A + \cos A)}

    =tan⁡4A= \tan 4A

  18. We have to prove, sin⁡(θ+ϕ)−2sin⁡θ+sin⁡(θ−ϕ)cos⁡(θ+ϕ)−2cos⁡θ+cos⁡(θ−ϕ)=tan⁡θ\frac{\sin (\theta + \phi) - 2\sin\theta + \sin (\theta - \phi)}{\cos (\theta + \phi) - 2\cos \theta + \cos(\theta - \phi)} = \tan\theta

    Pairing first and last term in both numerator and denominator, we get

    L.H.S. =2sin⁡θcos⁡ϕ+2sin⁡θ2cos⁡θcos⁡ϕ+2cos⁡θ= \frac{2\sin\theta\cos\phi + 2\sin\theta}{2\cos\theta\cos\phi + 2\cos\theta}

    =2sin⁡θ(cos⁡ϕ+1)2cos⁡θ(cos⁡ϕ+1)= \frac{2\sin\theta(\cos\phi + 1)}{2\cos\theta(\cos\phi + 1)}

    =tan⁡θ= \tan\theta

  19. We have to prove that, sin⁡A+2sin⁡3A+sin⁡5Asin⁡3A+2sin⁡5A+sin⁡7A=sin⁡3Asin⁡5A\frac{\sin A + 2\sin 3A + \sin 5A}{\sin 3A + 2\sin 5A + \sin 7A} = \frac{\sin 3A}{\sin 5A}

    Pairing first and last term in both numerator and denominator, we get

    L.H.S. =2sin⁡3Acos⁡2A+2sin⁡3A2sin⁡5Acos⁡2A+2sin⁡5A= \frac{2\sin3A\cos2A + 2\sin3A}{2\sin5A\cos2A + 2\sin5A}

    =sin⁡3A(cos⁡2A+1)sin⁡5A(cos⁡2A+1)= \frac{\sin3A(\cos 2A + 1)}{\sin5A(\cos 2A + 1)}

    =sin⁡3Asin⁡5A= \frac{\sin 3A}{\sin 5A}

  20. We have to prove that, sin⁡(A−C)+2sin⁡A+sin⁡(A+C)sin⁡(B−C)+2sin⁡B+sin⁡(B+C)=sin⁡Asin⁡B\frac{\sin(A - C) + 2\sin A + \sin(A + C)}{\sin (B - C) + 2\sin B + \sin(B + C)} = \frac{\sin A}{\sin B} Pairing first and last term in both numerator and denominator, we get

    L.H.S. =2sin⁡Acos⁡C+2sin⁡A2sin⁡Bcos⁡C+2sin⁡B= \frac{2\sin A\cos C + 2\sin A}{2\sin B\cos C + 2\sin B}

    =sin⁡A(cos⁡C+1)sin⁡B(cos⁡C+1)= \frac{\sin A(\cos C + 1)}{\sin B(\cos C + 1)}

    =sin⁡Asin⁡B= \frac{\sin A}{\sin B}

  21. We have to prove that, sin⁡A−sin⁡5A+sin⁡9A−sin⁡13Acos⁡A−cos⁡5A−cos⁡9A+cos⁡13A=cot⁡4A\frac{\sin A - \sin 5A + \sin 9A - \sin 13A}{\cos A - \cos 5A - \cos 9A + \cos 13 A} = \cot 4A

    Pairing first and last term and second and third term in both numerator and denominator, we get

    L.H.S. =−2cos⁡7Asin⁡6A+2cos⁡7Asin⁡2A3cos⁡7Acos⁡6A−2cos⁡7Acos⁡2A= \frac{-2\cos7A\sin6A + 2\cos7A\sin 2A}{3\cos7A\cos6A - 2\cos7A\cos2A}

    =2cos⁡7A(sin⁡2A−sin⁡6A)2cos⁡7A(cos⁡6A−cos⁡2A)= \frac{2\cos7A(\sin 2A - \sin 6A)}{2\cos 7A(\cos 6A - \cos 2A)}

    =−2cos⁡4Asin⁡2A−2sin⁡4Asin⁡2A= \frac{-2\cos 4A\sin 2A}{-2\sin 4A\sin 2A}

    =cot⁡4A= \cot 4A

  22. We have to prove that, sin⁡A+sin⁡Bsin⁡A−sin⁡B=tan⁡A+B2cot⁡A−B2\frac{\sin A + \sin B}{\sin A - \sin B} = \tan \frac{A + B}{2}\cot \frac{A - B}{2}

    L.H.S. =2sin⁡A+B2cos⁡A−B22cos⁡A+B2sin⁡A−B2= \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+B2cot⁡A−B2= \tan \frac{A + B}{2}\cot \frac{A - B}{2}

  23. We have to prove that, cos⁡A+cos⁡Bcos⁡B−cos⁡A=cot⁡A+B2cot⁡A−B2\frac{\cos A + \cos B}{\cos B - \cos A} = \cot \frac{A + B}{2}\cot \frac{A - B}{2}

    L.H.S. =2cos⁡A+B2cos⁡A−B22sin⁡A+B2sin⁡A−B2= \frac{2\cos\frac{A + B}{2}\cos\frac{A - B}{2}}{2\sin\frac{A + B}{2}\sin\frac{A - B}{2}}

    =cot⁡A+B2cot⁡A−B2= \cot\frac{A + B}{2}\cot\frac{A - B}{2}

  24. We have to prove that, sin⁡A+sin⁡Bcos⁡A+cos⁡B=tan⁡A+B2\frac{\sin A + \sin B}{\cos A + \cos B} = \tan \frac{A + B}{2}

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

    =tan⁡A+B2= \tan \frac{A + B}{2}

  25. We have to prove that, sin⁡A−sin⁡Bcos⁡B−cos⁡A=cot⁡A+B2\frac{\sin A - \sin B}{\cos B - \cos A} = \cot \frac{A + B}{2}

    L.H.S. =2cos⁡A+B2sin⁡A−B32sin⁡A+B2sin⁡A−B2= \frac{2\cos\frac{A + B}{2}\sin\frac{A - B}{3}}{2\sin\frac{A + B}{2}\sin\frac{A - B}{2}}

    =cot⁡A+B2= \cot \frac{A + B}{2}

  26. We have to prove that, cos⁡(A+B+C)+cos⁡(−A+B+C)+cos⁡(A−B+C)+cos⁡(A+B−C)sin⁡(A+B+C)+sin⁡(−A+B+C)−sin⁡(A−B+C)+sin⁡(A+B−C)=cot⁡B\frac{\cos(A + B + C) + \cos(-A + B + C) + \cos(A - B + C) + \cos(A + B - C)}{\sin(A + B + C)+\sin(-A + B + C) - \sin(A - B + C) + \sin(A + B - C)} = \cot B

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

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

    =2cos⁡Bcos⁡C2sin⁡Bcos⁡C=cot⁡B= \frac{2\cos B\cos C}{2\sin B\cos C} = \cot B

  27. We have to prove that, cos⁡3A+cos⁡5A+cos⁡7A+cos⁡15A=4cos⁡4Acos⁡5Acos⁡6A\cos 3A + \cos 5A + \cos 7A + \cos 15A = 4 \cos 4A\cos 5A \cos 6A

    Adding first and last and two middle terms together, we gte

    L.H.S. =2cos⁡9Acos⁡6A+2cos⁡6Acos⁡A= 2\cos9A\cos6A + 2\cos6A\cos A

    =2cos⁡6A(cos⁡9A+cos⁡A)= 2\cos6A(\cos9A + \cos A)

    =4cos⁡4Acos⁡5Acos⁡6A= 4\cos 4A \cos 5A \cos 6A

  28. We have to prove that, cos⁡(−A+B+C)+cos⁡(A−B+C)+cos⁡(A+B−C)+cos⁡(A+B+C)=4cos⁡Acos⁡Bcos⁡C\cos(-A + B + C) + \cos(A - B + C) + \cos(A + B - C) + \cos(A + B + C) = 4\cos A\cos B\cos C

    Adding first two and last two, we get

    L.H.S. =2cos⁡Ccos⁡(B−A)+2cos⁡(A+B)cos⁡C= 2\cos C\cos (B - A) + 2\cos (A + B)\cos C

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

    =4cos⁡Acos⁡Bcos⁡C= 4\cos A\cos B\cos C

  29. We have to prove that, sin⁡50∘−sin⁡70∘+sin⁡10∘=0\sin 50^\circ - \sin 70^\circ + \sin 10^\circ = 0

    L.H.S. =−2cos⁡60∘sin⁡10∘+sin⁡10∘= -2\cos 60^\circ \sin 10^\circ + \sin 10^\circ

    =−sin⁡10∘+sin⁡10∘=0= -\sin10^\circ + \sin10^\circ = 0

  30. We have to prove that, sin⁡10∘+sin⁡20∘+sin⁡40∘+sin⁡50∘=sin⁡70∘+sin⁡80∘\sin 10^\circ + \sin 20^\circ + \sin 40^\circ + \sin 50^\circ = \sin 70^\circ + \sin 80^\circ

    L.H.S. =sin⁡10∘+sin⁡50∘+sin⁡20∘+sin⁡40∘= \sin 10^\circ + \sin 50^\circ + \sin 20^\circ + \sin 40^\circ

    =2sin⁡30∘cos⁡20∘+2sin⁡30∘cos⁡10∘= 2\sin 30^\circ \cos 20^\circ + 2\sin 30^\circ \cos 10^\circ

    =2sin⁡30∘(cos⁡20∘+cos⁡10∘)= 2\sin30^\circ(\cos 20^\circ + \cos 10^\circ)

    =sin⁡70∘+sin⁡80∘[∵cos⁡θ=sin⁡(90∘−θ)]= \sin 70^\circ + \sin 80^\circ[\because \cos\theta = \sin(90^\circ - \theta)]

  31. We have to prove that, sin⁡α+sin⁡2α+sin⁡4α+sin⁡5α=4cos⁡α2cos⁡3α2sin⁡3α\sin\alpha + \sin 2\alpha + \sin 4\alpha + \sin 5\alpha = 4\cos \frac{\alpha}{2}\cos \frac{3\alpha}{2}\sin 3\alpha

    L.H.S. =sin⁡α+sin⁡5α+sin⁡2α+sin⁡4α= \sin\alpha + \sin 5\alpha + \sin2\alpha + \sin4\alpha

    =2sin⁡3αcos⁡2α+2sin⁡3αcos⁡α= 2\sin3\alpha\cos2\alpha + 2\sin3\alpha\cos\alpha

    =2sin⁡3α(cos⁡2α+cos⁡α)= 2\sin3\alpha(\cos2\alpha + \cos\alpha)

    =4cos⁡α2cos⁡3α2sin⁡3α= 4\cos \frac{\alpha}{2}\cos \frac{3\alpha}{2}\sin 3\alpha

  32. Given, cos⁡[θ+(n−32)ϕ]−cos⁡[θ+(n+32)ϕ]\cos\left[\theta + \left(n - \frac{3}{2}\right)\phi\right] - \cos\left[\theta + \left(n + \frac{3}{2}\right)\phi\right]

    =2sin⁡[θ+nϕ]sin⁡[ϕ2]= 2\sin\left[\theta + n\phi\right]\sin\left[\frac{\phi}{2}\right]

  33. Given, sin⁡[θ+(n−32)ϕ]+sin⁡[θ+(n+32)ϕ]\sin\left[\theta + \left(n - \frac{3}{2}\right)\phi\right] + \sin\left[\theta + \left(n + \frac{3}{2}\right)\phi\right]

    =2sin⁡[θ+nϕ]cos⁡[ϕ2]= 2\sin\left[\theta + n\phi\right]\cos\left[\frac{\phi}{2}\right]

  34. Given, 2sin⁡5θsin⁡7θ2\sin5\theta\sin7\theta

    Let the angles are AA and BB then cos⁡C−cos⁡D=2sin⁡C+D2cos⁡D−C2\cos C - \cos D = 2\sin \frac{C + D}{2}\cos\frac{D - C}{2}

    Thus, comparing, we get

    C+D=14,D−C=10C + D = 14, D - C = 10

    D=12,C=2D = 12, C = 2

    Thus, required expression is cos⁡2θ−cos⁡12θ\cos 2\theta - \cos12\theta

  35. Given, 2cos⁡7θsin⁡5θ2\cos7\theta\sin5\theta

    =sin⁡12θ+sin⁡2θ= \sin 12\theta + \sin 2\theta

  36. Given, 2cos⁡11θcos⁡3θ2\cos 11\theta\cos 3\theta

    =cos⁡14θ+cos⁡8θ= \cos 14\theta + \cos 8\theta

  37. Given, 2sin⁡54∘sin⁡66∘2\sin54^\circ\sin66^\circ

    =cos⁡12∘−cos⁡120∘=\cos12^\circ - \cos120^\circ

  38. We have to prove that sin⁡θ2sin⁡7θ2+sin⁡3θ2sin⁡11θ2=sin⁡2θsin⁡5θ\sin\frac{\theta}{2}\sin\frac{7\theta}{2} + \sin \frac{3\theta}{2}\sin\frac{11\theta}{2} =\sin 2\theta\sin 5\theta

    L.H.S. =12(cos⁡3θ−cos⁡4θ)+12(cos⁡4θ−cos⁡7θ)= \frac{1}{2}(\cos 3\theta - \cos 4\theta) + \frac{1}{2}(\cos 4\theta - \cos 7\theta)

    =sin⁡2θ+sin⁡5θ= \sin2\theta + \sin5\theta

  39. We have to prove that, cos⁡2θcos⁡θ2−cos⁡3θcos⁡9θ2=sin⁡5θsin⁡5θ2\cos 2\theta\cos \frac{\theta}{2} -\cos3\theta\cos\frac{9\theta}{2} = \sin5\theta\sin\frac{5\theta}{2}

    L.H.S. =12(cos⁡5θ2+cos⁡3θ2)−12(cos⁡15θ2+cos⁡3θ2)= \frac{1}{2}\left(\cos\frac{5\theta}{2} + \cos \frac{3\theta}{2}\right) - \frac{1}{2}\left(\cos \frac{15\theta}{2} + \cos \frac{3\theta}{2}\right)

    =12(2sin⁡5θsin⁡5θ2)= \frac{1}{2}\left(2\sin 5\theta \sin \frac{5\theta}{2}\right)

    =sin⁡5θsin⁡5θ2= \sin5\theta\sin\frac{5\theta}{2}

  40. We have to prove that, sin⁡Asin⁡(A+2B)−sin⁡Bsin⁡(B+2A)=sin⁡(A−B)sin⁡(A+B)\sin A\sin(A + 2B) - \sin B\sin(B + 2A) = \sin(A - B)\sin(A + B)

    L.H.S. =12(2sin⁡Asin⁡(A+2B)−2sin⁡Bsin⁡(B+2A))= \frac{1}{2}(2\sin A\sin(A + 2B) - 2\sin B\sin(B + 2A))

    =12(cos⁡B−cos⁡(A+B)−cos⁡A−cos⁡(A+B))= \frac{1}{2}(\cos B - \cos(A + B) - \cos A - \cos (A + B))

    =122sin⁡(A−B)sin⁡(A+B)= \frac{1}{2}2\sin(A - B)\sin(A + B)

  41. We have to prove that, (sin⁡3A+sin⁡A)sin⁡A+(cos⁡3A−cos⁡A)cos⁡A=0(\sin 3A + \sin A)\sin A + (\cos 3A - \cos A)\cos A = 0

    L.H.S. =2sin⁡2Acos⁡Asin⁡A−2sin⁡2Asin⁡Acos⁡A=0= 2\sin2A\cos A\sin A - 2\sin2A\sin A\cos A = 0

  42. We have to prove that, 2sin⁡(A−C)cos⁡C−sin⁡(A−2C)2sin⁡(B−C)cos⁡C−sin⁡(B−2C)=sin⁡Asin⁡B\frac{2\sin(A - C)\cos C - \sin(A - 2C)}{2\sin(B - C)\cos C - \sin(B - 2C)} = \frac{\sin A}{\sin B}

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

    =sin⁡Asin⁡B= \frac{\sin A}{\sin B}

  43. We have to prove that, sin⁡Asin⁡2A+sin⁡3Asin⁡6A+sin⁡4Asin⁡13Asin⁡Acos⁡2A+sin⁡3Acos⁡6A+sin⁡4Acos⁡13A=tan⁡9A\frac{\sin A\sin 2A + \sin 3A\sin 6A + \sin4A\sin 13A}{\sin A\cos2A + \sin 3A\cos 6A + \sin 4A\cos 13A} = \tan 9A

    L.H.S. =cos⁡A−cos⁡3A+cos⁡3A−cos⁡9A+cos⁡9A−cos⁡17Asin⁡3A−sin⁡A+sin⁡9A−sin⁡3A+sin⁡17A−sin⁡9A= \frac{\cos A - \cos 3A + \cos 3A - \cos 9A + \cos 9A - \cos 17A}{\sin 3A - \sin A + \sin 9A - \sin 3A + \sin 17A - \sin 9A}

    =cos⁡A−cos⁡17Asin⁡17A−sin⁡A= \frac{\cos A - \cos 17A}{\sin 17A - \sin A}

    =2sin⁡8Asin⁡9A2cos⁡9Asin⁡8A=tan⁡9A= \frac{2\sin 8A\sin 9A}{2\cos 9A\sin 8A} = \tan 9A

  44. We have to prove that, cos⁡2Acos⁡3A−cos⁡2Acos⁡7A+cos⁡Acos⁡10Asin⁡4Asin⁡3A−sin⁡2Asin⁡5A+sin⁡4Asin⁡7A=cot⁡6Acot⁡5A\frac{\cos 2A\cos 3A - \cos 2A\cos 7A + \cos A\cos 10A}{\sin 4A\sin 3A - \sin 2A\sin 5A + \sin 4A\sin 7A} =\cot 6A\cot 5A

    L.H.S. =cos⁡5A+cos⁡A−cos⁡9A−cos⁡5A+cos⁡11A+cos⁡9Acos⁡A−cos⁡7A−cos⁡3A+cos7A+cos⁡3A−cos⁡11A= \frac{\cos 5A + \cos A - \cos 9A - \cos 5A + \cos 11A + \cos 9A}{\cos A - \cos 7A - \cos 3A + cos 7A + \cos 3A - \cos 11A}

    =cos⁡A+cos⁡11Acos⁡A−cos⁡11A= \frac{\cos A + \cos 11A}{\cos A - \cos 11A}

    =cos⁡6Acos⁡5Asin⁡6Asin⁡5A=cot⁡6Acot⁡5A= \frac{\cos 6A\cos 5A}{\sin 6A\sin 5A} = \cot 6A\cot 5A

  45. We have to prove that, cos⁡(36∘−A)cos⁡(36∘+A)+cos⁡(54∘+A)cos⁡(54∘−A)=cos⁡2A\cos(36^\circ - A)\cos(36^\circ + A) + \cos(54^\circ + A)\cos(54^\circ - A) = \cos 2A

    L.H.S. =12(cos⁡(72∘)+cos⁡2A)+12(cos⁡108∘+cos⁡2A)= \frac{1}{2}(\cos (72^\circ)+ \cos 2A) + \frac{1}{2}(\cos 108^\circ + \cos 2A)

    =12(sin⁡18∘+−sin⁡18∘+2cos⁡2A)[∵sin⁡18∘=cos⁡(90∘−18∘)=cos⁡72∘= \frac{1}{2}(\sin 18^\circ + - \sin 18^\circ + 2\cos 2A)[\because \sin 18^\circ = \cos(90^\circ - 18^\circ) = \cos 72^\circ and cos⁡108∘=cos⁡(90∘+18∘)=−sin⁡18∘]\cos 108^\circ = \cos(90^\circ + 18^\circ) = -\sin 18^\circ]

    =cos⁡2A= \cos 2A

  46. We have to prove that cos⁡Asin⁡(B−C)+cos⁡Bsin⁡(C−A)+cos⁡Csin⁡(A−B)=0\cos A\sin(B - C) + \cos B\sin(C - A) + \cos C\sin(A - B) = 0

    L.H.S. =12[sin⁡(A+B−C)−sin⁡(A−B+C)+sin⁡(B+C−A)−sin⁡(B−C+A)+sin⁡(A−B+C)−sin⁡(C−A+B)]=0= \frac{1}{2}[\sin(A + B - C) - \sin(A - B + C) + \sin(B + C - A) - \sin (B - C + A) + \sin(A - B + C) - \sin(C - A + B)] = 0

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

    L.H.S. =12(2sin⁡(45∘+A)sin⁡(45∘−A))=12[cos⁡2A−cos⁡90∘]=12cos⁡2A= \frac{1}{2}(2\sin(45^\circ + A)\sin(45^\circ - A)) = \frac{1}{2}[\cos 2A - \cos 90^\circ] = \frac{1}{2}\cos 2A

  48. We have to prove that, sin⁡(β−γ)cos⁡(α−δ)+sin⁡(γ−α)cos⁡(β−δ)+sin⁡(α−β)cos⁡(γ−δ)=0\sin(\beta - \gamma)\cos(\alpha - \delta) + \sin(\gamma - \alpha)\cos(\beta - \delta) + \sin(\alpha - \beta)\cos(\gamma - \delta) = 0

    L.H.S. =12[sin⁡(α+β−γ−δ)+sin⁡(β+δ−γ−α)+sin⁡(γ−α+β−δ)+sin⁡(γ−α−β+δ)+sin⁡(α−β+γ−δ)−sin⁡(α−β−γ+δ)]= \frac{1}{2}[\sin(\alpha + \beta - \gamma - \delta) + \sin(\beta + \delta - \gamma - \alpha) + \sin(\gamma -\alpha + \beta - \delta) + \sin(\gamma - \alpha - \beta + \delta) + \sin(\alpha - \beta + \gamma - \delta) - \sin(\alpha - \beta - \gamma + \delta)]

    =0= 0

  49. We have to prove that, 2cos⁡π13cos⁡9π13+cos⁡3π13+cos⁡5π13=02\cos\frac{\pi}{13}\cos \frac{9\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0

    L.H.S. =cos⁡10π13+cos⁡8π13+cos⁡(π−10π13)+cos⁡(π−8π13)=\cos \frac{10\pi}{13} + \cos \frac{8\pi}{13} + \cos \left(\pi - \frac{10\pi}{13}\right) + \cos \left(\pi - \frac{8\pi}{13}\right)

    =cos⁡10π13+cos⁡8π13−cos⁡10π13−cos⁡8π13=0= \cos \frac{10\pi}{13} + \cos \frac{8\pi}{13} - \cos \frac{10\pi}{13} - \cos \frac{8\pi}{13} = 0

  50. We have to prove that, cos⁡55∘+cos⁡65∘+cos⁡175∘=0\cos 55^\circ + \cos65^\circ + \cos 175^\circ = 0

    L.H.S. =2cos⁡60∘cos⁡5∘+cos⁡(180∘−5∘)= 2\cos 60^\circ\cos5^\circ + \cos(180^\circ - 5^\circ)

    =2.12.cos⁡5∘−cos⁡5∘=0= 2.\frac{1}{2}.\cos5^\circ - \cos 5^\circ = 0

  51. We have to prove that, cos⁡18∘−sin⁡18∘=2sin⁡27∘\cos 18^\circ -\sin 18^\circ = \sqrt{2}\sin 27^\circ

    L.H.S. =cos⁡18∘−cos⁡(90∘−72∘)=cos⁡18∘−cos⁡72∘= \cos18^\circ - \cos(90^\circ - 72^\circ) = \cos 18^\circ - \cos 72^\circ

    =2sin⁡45∘sin⁡27∘=2sin⁡27∘= 2\sin 45^\circ\sin 27^\circ = \sqrt{2}\sin27^\circ

  52. We have to prove that, sin⁡A+sin⁡2A+sin⁡4A+sin⁡5Acos⁡A+cos⁡2A+cos⁡4A+cos⁡5A=tan⁡3A\frac{\sin A + \sin 2A + \sin 4A + \sin 5A}{\cos A + \cos 2A + \cos 4A + \cos 5A} = \tan 3A

    L.H.S. =(sin⁡5A+sin⁡A)+(sin⁡4A+sin⁡2A)(cos⁡5A+cos⁡A)+(cos⁡4A+cos⁡2A)= \frac{(\sin 5A + \sin A) + (\sin 4A + \sin 2A)}{(\cos 5A + \cos A) + (\cos 4A + \cos 2A)}

    =2sin⁡3Acos⁡2A+2sin⁡3Acos⁡A2cos⁡3Acos⁡2A+2cos⁡3Acos⁡A= \frac{2\sin 3A\cos 2A + 2\sin 3A\cos A}{2\cos3A\cos2A + 2\cos 3A\cos A}

    =sin⁡3A(cos⁡2A+cos⁡A)cos⁡3A(cos⁡2A+cos⁡A)=tan⁡3A= \frac{\sin 3A(\cos 2A + \cos A)}{\cos 3A(\cos 2A + \cos A)} = \tan 3A

  53. L.H.S =(2cos⁡A+B2cos⁡A−B22cos⁡A+B2sin⁡A−B2)n+(2sin⁡A+B2cos⁡A−B22sin⁡A+B2sin⁡B−A2)n= \left(\frac{2\cos \frac{A + B}{2}\cos \frac{A - B}{2}}{2\cos\frac{A + B}{2}\sin\frac{A - B}{2}}\right)^n + \left(\frac{2\sin\frac{A + B}{2}\cos\frac{A - B}{2}}{2\sin\frac{A + B}{2}\sin\frac{B - A}{2}}\right)^n

    =(cos⁡A−B2)n+(−cot⁡A−B2)n= \left(\cos\frac{A - B}{2}\right)^n + \left(-\cot\frac{A - B}{2}\right)^n

    =cot⁡nA−B2[1+(−1)n]=\cot^n\frac{A - B}{2}[1 + (-1)^n] which is 00 if nn is odd and 2cos⁡nA−B22\cos^n\frac{A - B}{2} if nn is even.

  54. Given α,β,γ\alpha, \beta, \gamma are in A.P. ∴2β=α+γ\therefore 2\beta = \alpha + \gamma

    R.H.S. =sin⁡α−sin⁡γcos⁡γ−cos⁡α=2cos⁡α+γ2sin⁡α−γ22sin⁡γ+α2sin⁡α−γ2= \frac{\sin\alpha - \sin\gamma}{\cos\gamma - \cos\alpha} = \frac{2\cos \frac{\alpha + \gamma}{2}\sin\frac{\alpha - \gamma}{2}}{2\sin\frac{\gamma + \alpha}{2}\sin\frac{\alpha - \gamma}{2}}

    =cot⁡α+γ2=cot⁡β= \cot\frac{\alpha + \gamma}{2} = \cot\beta

  55. Given sin⁡θ+sin⁡ϕ=3(cos⁡ϕ−cos⁡θ)\sin\theta + \sin\phi = \sqrt{3}(\cos\phi - \cos\theta)

    2sin⁡θ+ϕ2cos⁡θ−ϕ2=3.2.sin⁡θ+ϕ2sin⁡θ−ϕ22\sin\frac{\theta + \phi}{2}\cos\frac{\theta - \phi}{2} = \sqrt{3}.2.\sin\frac{\theta + \phi}{2}\sin\frac{\theta - \phi}{2}

    =sin⁡θ+ϕ2[cos⁡θ−ϕ2−3sin⁡θ−ϕ]=0= \sin\frac{\theta + \phi}{2}\left[\cos\frac{\theta - \phi}{2} - \sqrt{3}\sin\frac{\theta - \phi}{}\right] = 0

    ∴sin⁡θ+ϕ2=0\therefore \sin\frac{\theta + \phi}{2} = 0 or tan⁡θ−ϕ2=13\tan\frac{\theta - \phi}{2} = \frac{1}{\sqrt{3}}

    ∴θ+ϕ=0∘\therefore \theta + \phi = 0^\circ or θ−ϕ=60∘\theta - \phi = 60^\circ

    Now, sin⁡3θ+sin⁡3ϕ=2sin⁡3(θ+ϕ)2sin⁡3(θ−ϕ)2=0\sin3\theta + \sin3\phi = 2\sin\frac{3(\theta + \phi)}{2}\sin\frac{3(\theta - \phi)}{2} = 0

    [∵\because when θ+ϕ=0;sin⁡3(θ+ϕ)2=0\theta + \phi = 0; \sin\frac{3(\theta + \phi)}{2} = 0 and when θ−ϕ=60∘;cos⁡3(θ−ϕ)2=0\theta - \phi = 60^\circ; \cos\frac{3(\theta - \phi)}{2} = 0 ]

  56. We have to prove that sin⁡65∘+cos65∘=2cos⁡20∘\sin 65^\circ + cos 65^\circ = \sqrt{2}\cos 20^\circ

    L.H.S. =cos⁡(90∘−65∘)+cos⁡65∘=cos⁡25∘cos⁡65∘= \cos(90^\circ - 65^\circ) + \cos 65^\circ = \cos 25^\circ\cos65^\circ

    =2cos⁡45∘cos⁡20∘=2.12cos⁡20∘=2cos⁡20∘= 2\cos 45^\circ\cos20^\circ = 2.\frac{1}{\sqrt{2}}\cos20^\circ = \sqrt{2}\cos20^\circ

  57. We have to prove that sin⁡47∘+cos⁡77∘=cos⁡17∘\sin 47^\circ + \cos 77^\circ = \cos 17^\circ

    L.H.S. =cos⁡(90∘−47∘)+cos⁡77∘= \cos(90^\circ - 47^\circ) + \cos 77^\circ

    =cos⁡43∘+cos⁡77∘= \cos 43^\circ + \cos77^\circ

    =2cos⁡60∘cos⁡17∘=2\cos 60^\circ \cos 17^\circ

    =cos⁡17∘[∵2cos⁡60∘=2.12=1= \cos 17^\circ[\because 2\cos60^\circ = 2.\frac{1}{2} = 1

  58. We have to prove that, cos⁡10∘−sin⁡10∘cos⁡10∘+sin⁡10∘=tan⁡35∘\frac{\cos 10^\circ - \sin 10^\circ}{\cos 10^\circ + \sin 10^\circ} = \tan 35^\circ

    L.H.S. =cos⁡10∘−cos⁡(90∘−10∘)cos⁡10∘+cos⁡(90∘−10∘)= \frac{\cos10^\circ - \cos(90^\circ - 10^\circ)}{\cos10^\circ + \cos(90^\circ - 10^\circ)}

    =cos⁡10∘−cos⁡80∘cos⁡10∘+cos⁡80∘= \frac{\cos10^\circ - \cos80^\circ}{\cos10^\circ + \cos80^\circ}

    =2cos⁡45∘cos⁡35∘2cos⁡45∘cos⁡35∘=tan⁡35∘= \frac{2\cos45^\circ\cos35^\circ}{2\cos45^\circ\cos35^\circ} = \tan 35^\circ

  59. We have to prove that, cos⁡80∘+cos⁡40∘−cos20∘=0\cos 80^\circ + \cos 40^\circ - cos 20^\circ = 0

    L.H.S. =2cos⁡60∘cos⁡20∘−cos⁡20∘= 2\cos60^\circ\cos20^\circ - \cos20^\circ

    =cos⁡20∘−cos⁡20∘[∵2cos⁡60∘=2.12=1]= \cos20^\circ - \cos20^\circ[\because 2\cos60^\circ = 2.\frac{1}{2} = 1]

  60. We have to prove that cos⁡π5+cos⁡2π5+cos⁡6π5+cos⁡7π5=0\cos\frac{\pi}{5} + \cos \frac{2\pi}{5} + \cos\frac{6\pi}{5} + \cos \frac{7\pi}{5} = 0

    L.H.S. =cos⁡7π5+cos⁡π5+cos⁡2π5+cos⁡6π5= \cos\frac{7\pi}{5} + \cos\frac{\pi}{5} + \cos\frac{2\pi}{5} + \cos\frac{6\pi}{5}

    =2cos⁡4π5cos⁡3π5+2cos⁡4π5cos⁡2π5= 2\cos\frac{4\pi}{5}\cos\frac{3\pi}{5} + 2\cos\frac{4\pi}{5}\cos\frac{2\pi}{5}

    =2cos⁡4π5[cos⁡3π5+cos⁡(π−3π5)]= 2\cos\frac{4\pi}{5}\left[\cos\frac{3\pi}{5} + \cos \left(\pi - \frac{3\pi}{5}\right)\right]

    =2cos⁡4π5(cos⁡3π5−cos⁡3π5)=0= 2\cos\frac{4\pi}{5}\left(\cos\frac{3\pi}{5} - \cos \frac{3\pi}{5}\right) = 0

  61. We have to prove that cos⁡α+cos⁡β+cos⁡γ+cos⁡(α+β+γ)=4cos⁡α+β2cos⁡β+γ2cos⁡γ+α2\cos\alpha + \cos\beta + \cos\gamma + \cos(\alpha + \beta + \gamma) = 4\cos\frac{\alpha + \beta}{2}\cos\frac{\beta + \gamma}{2}\cos \frac{\gamma + \alpha}{2}

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

    =2cos⁡(α+β+γ2)cos⁡β+γ2+2cos⁡β+γ2cos⁡β−γ2= 2\cos\left(\alpha + \frac{\beta + \gamma}{2}\right)\cos\frac{\beta + \gamma}{2} + 2\cos\frac{\beta+\gamma}{2}\cos\frac{\beta - \gamma}{2}

    =2cos⁡β+γ2[2cos⁡(α+β+γ2)+cos⁡β−γ2]= 2\cos\frac{\beta + \gamma}{2}\left[2\cos\left(\alpha +\frac{\beta + \gamma}{2}\right) + \cos\frac{\beta - \gamma}{2}\right]

    =4cos⁡α+β2cos⁡β+γ2cos⁡γ+α2=4\cos\frac{\alpha + \beta}{2}\cos\frac{\beta + \gamma}{2}\cos \frac{\gamma + \alpha}{2}

  62. Given, sin⁡α−sin⁡β=13\sin\alpha - \sin\beta = \frac{1}{3} and cosbeta - cosalpha = frac{1}{2}

    Dividing we get, sin⁡α−sin⁡βcos⁡β−cos⁡α=23\frac{\sin\alpha - \sin\beta}{\cos\beta - \cos\alpha} = \frac{2}{3}

    ⇒2cos⁡α+β2sin⁡α−β22sin⁡α+β2sin⁡α−β2=23\Rightarrow \frac{2\cos\frac{\alpha + \beta}{2}\sin\frac{\alpha - \beta}{2}}{2\sin\frac{\alpha + \beta}{2}\sin\frac{\alpha - \beta}{2}} = \frac{2}{3}

    ⇒cot⁡α+β2=23\Rightarrow \cot\frac{\alpha + \beta}{2} = \frac{2}{3}

  63. Given, cosec⁡A+secA=cosec⁡B+sec⁡B\cosec A + sec A = \cosec B + \sec B

    sec⁡A−sec⁡B=cosec⁡B−cosec⁡A\sec A - \sec B = \cosec B - \cosec A

    ⇒cos⁡B−cos⁡Acos⁡Acos⁡B=sin⁡A−sin⁡Bsin⁡Asin⁡B\Rightarrow \frac{\cos B - \cos A}{\cos A\cos B} = \frac{\sin A - \sin B}{\sin A\sin B}

    ⇒2sin⁡A+B2sin⁡A−B2cos⁡Acos⁡B=2cos⁡A+B2sin⁡A−B2sin⁡Asin⁡B\Rightarrow \frac{2\sin\frac{A + B}{2}\sin \frac{A - B}{2}}{\cos A\cos B} = \frac{2\cos\frac{A + B}{2}\sin\frac{A - B}{2}}{\sin A\sin B}

    ⇒tan⁡Atan⁡B=cot⁡A+B2\Rightarrow \tan A\tan B = \cot \frac{A + B}{2}

  64. Given, sec⁡(θ+α)+sec⁡(θ−α)=2sec⁡θ\sec(\theta + \alpha) + \sec(\theta - \alpha) = 2\sec\theta

    ⇒1cos⁡(θ+α)+1cos⁡(θ−α)=2cos⁡θ\Rightarrow \frac{1}{\cos(\theta + \alpha)} + \frac{1}{\cos(\theta - \alpha)} = \frac{2}{\cos\theta}

    cos⁡θ[cos⁡(θ−α)+cos⁡(θ+α)]=2cos⁡(θ+α)cos⁡(θ−α)\cos\theta[\cos(\theta - \alpha) + \cos(\theta + \alpha)] = 2\cos(\theta + \alpha)\cos(\theta - \alpha)

    cos⁡θ.2cos⁡θcos⁡α=cos⁡2θ+cos⁡2α\cos\theta.2\cos\theta\cos\alpha = \cos2\theta + \cos2\alpha

    We know that [cos(theta + theta) = costheta.costheta - sinthetasintheta = 2cos^2theta - 1]

    Thus, the above equation becomes

    2cos⁡2θcos⁡α=2cos⁡2θ−1+2cos⁡2α−12\cos^2\theta\cos\alpha = 2\cos^2\theta - 1 + 2\cos^2\alpha - 1

    2cos⁡2θ(cos⁡α−1)=2(cos⁡2α−1)2\cos^2\theta(\cos\alpha - 1) = 2(\cos^2\alpha - 1)

    ⇒cos⁡2θ=1+cos⁡α\Rightarrow \cos^2\theta = 1 + \cos\alpha

  65. We have to prove that sin⁡50∘cos⁡85∘=1−2sin⁡35∘22\sin50^\circ\cos85^\circ = \frac{1 - \sqrt{2}\sin 35^\circ}{2\sqrt{2}}

    L.H.S. =12[sin⁡(85∘+sin⁡50∘)−sin⁡(85∘−50∘)]= \frac{1}{2}[\sin(85^\circ + \sin50^\circ) - \sin(85^\circ - 50^\circ)]

    =12[sin⁡135∘−sin⁡35∘]= \frac{1}{2}[\sin135^\circ - \sin35^\circ]

    =12[12−sin⁡35∘]= \frac{1}{2}\left[\frac{1}{\sqrt{2}} - \sin35^\circ\right]

    =1−2sin⁡35∘22= \frac{1 - \sqrt{2}\sin 35^\circ}{2\sqrt{2}}

  66. We have to prove that, sin⁡20∘sin⁡40∘sin⁡80∘=38\sin 20^\circ \sin 40^\circ\sin 80^\circ = \frac{\sqrt{3}}{8}

    L.H.S. =12(2sin⁡80∘sin⁡40∘)sin⁡20∘= \frac{1}{2}(2\sin80^\circ\sin40^\circ)\sin20^\circ

    =12[cos⁡(80∘−40∘)−cos⁡(80∘+40∘)]sin⁡20∘= \frac{1}{2}[\cos(80^\circ - 40^\circ) - \cos(80^\circ + 40^\circ)]\sin20^\circ

    =12(cos⁡40∘−cos⁡120∘)sin⁡206∘= \frac{1}{2}(\cos40^\circ - \cos120^\circ)\sin 206\circ

    =14(2cos⁡40∘sin⁡20∘−2cos⁡120∘sin⁡20∘)= \frac{1}{4}(2\cos40^\circ\sin20^\circ - 2\cos120^\circ\sin20^\circ)

    =14[sin⁡(40∘+20∘)−sin⁡(40∘−20∘)−2.−12sin⁡20∘]= \frac{1}{4}[\sin(40^\circ + 20^\circ) - \sin(40^\circ - 20^\circ) - 2.-\frac{1}{2}\sin20^\circ]

    =14[sin⁡60∘−sin⁡20∘+sin⁡20∘]=14sin⁡60∘=38= \frac{1}{4}[\sin60^\circ - \sin20^\circ + \sin20^\circ] = \frac{1}{4}\sin60^\circ = \frac{\sqrt{3}}{8}

  67. 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. =12sin⁡A[2sin⁡(60∘−A)sin⁡(60∘+A)]= \frac{1}{2}\sin A[2\sin(60^\circ - A)\sin(60^\circ + A)]

    =12sin⁡A[cos⁡(60∘+A−60∘+A)−cos⁡(60∘+A+60∘−A)]= \frac{1}{2}\sin A[\cos(60^\circ + A - 60^\circ + A) - \cos(60^\circ + A + 60^\circ - A)]

    =12sin⁡A(cos⁡2A−cos⁡120∘)= \frac{1}{2}\sin A(\cos 2A - \cos 120^\circ)

    =14(2sin⁡Acos⁡2A−2cos⁡120∘sin⁡A)= \frac{1}{4}(2\sin A\cos 2A - 2\cos 120^\circ \sin A)

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

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

  68. Let p=sin⁡αsin⁡βp = \sin\alpha\sin\beta

    =122sin⁡αsin⁡β= \frac{1}{2}2\sin\alpha\sin\beta

    =12[cos⁡(α−β)−cos⁡(α+beta)]= \frac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha + beta)]

    =12[cos⁡(α−β)−cos⁡90∘][∵α+β=90∘ (given)]= \frac{1}{2}[\cos(\alpha - \beta) - \cos90^\circ][\because \alpha + \beta = 90^\circ~\text{(given)}]

    =12cos⁡(α−β)= \frac{1}{2}\cos(\alpha - \beta)

    Maximum value of cos⁡(α−β)\cos(\alpha - \beta) is 1,1, hence maximum value of pp is 12\frac{1}{2}

  69. We have to prove that, sin⁡25∘cos⁡115∘=12(sin⁡40∘−1)\sin 25^\circ\cos 115^\circ = \frac{1}{2}(\sin 40^\circ - 1)

    L.H.S. =sin⁡25∘cos⁡115∘= \sin 25^\circ\cos 115^\circ

    =122sin⁡25∘cos⁡115∘=12[sin⁡140∘−sin⁡90∘]= \frac{1}{2}2\sin25^\circ\cos115^\circ = \frac{1}{2}[\sin140^\circ - \sin90^\circ]

    =12[cos⁡50∘−1]=12[sin⁡40∘−1]= \frac{1}{2}[\cos50^\circ - 1] = \frac{1}{2}[\sin40^\circ - 1]

  70. We have to prove that, sin⁡20∘sin⁡40∘sin⁡60∘sin⁡80∘=316\sin 20^\circ \sin 40^\circ\sin 60^\circ \sin80^\circ = \frac{3}{16}

    L.H.S. =12[(2sin⁡20∘sin⁡80∘)(sin⁡40∘sin⁡60∘)]= \frac{1}{2}[(2\sin 20^\circ\sin80^\circ)(\sin40^\circ\sin60^\circ)]

    =12(cos⁡60∘−cos⁡100∘)32sin⁡40∘= \frac{1}{2}(\cos60^\circ - \cos100^\circ)\frac{\sqrt{3}}{2}\sin40^\circ

    =34[cos⁡60∘sin⁡40∘+sin⁡10∘sin⁡40∘]= \frac{\sqrt{3}}{4}[\cos60^\circ\sin40^\circ + \sin10^\circ\sin40^\circ]

    =38[2cos⁡60∘sin⁡40∘+2cos⁡80∘sin⁡40∘]= \frac{\sqrt{3}}{8}[2\cos60^\circ\sin40^\circ + 2\cos80^\circ\sin40^\circ]

    =38[sin⁡100∘−sin⁡20∘+sin⁡120∘−sin⁡40∘]= \frac{\sqrt{3}}{8}[\sin100^\circ - \sin 20^\circ + \sin 120^\circ - \sin 40^\circ]

    =38[cos⁡10∘−(sin⁡20∘+sin⁡40∘)+cos⁡30∘]= \frac{\sqrt{3}}{8}[\cos10^\circ - (\sin20^\circ + \sin40^\circ) + \cos30^\circ]

    =38[cos⁡10∘−2sin⁡30∘cos⁡10∘+32]= \frac{\sqrt{3}}{8}[\cos10^\circ - 2\sin30^\circ\cos10^\circ + \frac{\sqrt{3}}{2}]

    =38[cos⁡10∘−2.12.cos⁡10∘+32]=316= \frac{3}{8}[\cos10^\circ - 2.\frac{1}{2}.\cos10^\circ + \frac{3}{2}] = \frac{3}{16}

  71. We have to prove that, cos⁡20∘cos⁡40∘cos⁡80∘=18\cos 20^\circ\cos40^\circ\cos80^\circ = \frac{1}{8}

    L.H.S. =12cos⁡20∘cos⁡80∘cos⁡40∘= \frac{1}{2\cos20^\circ\cos80^\circ}\cos40^\circ

    =12[cos⁡100∘+cos⁡60∘]cos⁡40∘= \frac{1}{2}[\cos100^\circ + \cos60^\circ]\cos40^\circ

    =12[−sin⁡10∘cos⁡40∘+12cos⁡40∘]= \frac{1}{2}[-\sin10^\circ\cos40^\circ + \frac{1}{2}\cos40^\circ]

    =14[−2cos⁡40∘sin⁡10∘+cos⁡40∘]= \frac{1}{4}[-2\cos40^\circ\sin10^\circ + \cos40^\circ]

    =14[−sin⁡50∘+sin30∘+cos⁡40∘]= \frac{1}{4}[-\sin50^\circ + sin30^\circ + \cos40^\circ]

    =18[∵sin⁡50∘=cos⁡40∘ and sin⁡30∘=12]= \frac{1}{8}[\because \sin50^\circ = \cos40^\circ~\text{and}~\sin30^\circ = \frac{1}{2}]

  72. We have to prove that, tan⁡20∘tan⁡40∘tan⁡60∘tan⁡80∘=3\tan20^\circ\tan40^\circ\tan60^\circ\tan80^\circ = 3

    Using results of 70 and 71 it can be solved.

  73. We have to prove that, cos⁡10∘cos⁡30∘cos⁡50∘cos⁡70∘=316\cos10^\circ\cos30^\circ\cos50^\circ\cos70^\circ = \frac{3}{16}

    L.H.S. =12(2cos⁡10∘cos⁡70∘)(32cos⁡50∘)= \frac{1}{2}(2\cos10^\circ\cos70^\circ)(\frac{\sqrt{3}}{2}\cos50^\circ)

    =34[cos⁡80∘+cos⁡60∘]cos⁡50∘= \frac{\sqrt{3}}{4}[\cos80^\circ + \cos60^\circ]\cos50^\circ

    =34[cos⁡80∘cos⁡50∘+12cos⁡50∘]= \frac{\sqrt{3}}{4}[\cos80^\circ\cos50^\circ + \frac{1}{2}\cos50^\circ]

    =38[2cos⁡80∘cos⁡50∘+cos⁡50∘]= \frac{\sqrt{3}}{8}[2\cos80^\circ\cos50^\circ + \cos50^\circ]

    =38[cos⁡130∘+cos⁡30∘+cos⁡50∘]= \frac{\sqrt{3}}{8}[\cos 130^\circ + \cos30^\circ + \cos50^\circ]

    =38[cos⁡(180∘−50∘)+cos⁡30∘+cos⁡50∘]= \frac{\sqrt{3}}{8}[\cos(180^\circ - 50^\circ) + \cos30^\circ + \cos50^\circ]

    =316= \frac{3}{16}

  74. We have to prove that, 4cos⁡θcos⁡(π3+θ)cos⁡(π3−θ)=cos⁡3θ4\cos\theta\cos\left(\frac{\pi}{3} + \theta\right)\cos\left(\frac{\pi}{3} - \theta\right) = \cos3\theta

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

    =2cos⁡θ[cos⁡(2π3)+cos⁡2θ]= 2\cos\theta\left[\cos\left(\frac{2\pi}{3}\right) + \cos2\theta\right]

    =−cos⁡θ+2cos⁡θcos⁡2θ=−cos⁡θ+cos⁡3θ+cos⁡θ=cos⁡3θ= -\cos\theta + 2\cos\theta\cos2\theta = -\cos\theta + \cos3\theta + \cos\theta = \cos3\theta

  75. We have to prove that tan⁡θtan⁡(60∘−θ)tan⁡(60∘+θ)=tan⁡3θ\tan\theta\tan(60^\circ - \theta)\tan(60^\circ + \theta) = \tan3\theta

    We have just proven, 4cos⁡θcos⁡(π3+θ)cos⁡(π3−θ)=cos⁡3θ4\cos\theta\cos\left(\frac{\pi}{3} + \theta\right)\cos\left(\frac{\pi}{3} - \theta\right) = \cos3\theta

    Let us evaluate sin⁡θsin⁡(60∘−θ)sin⁡(60∘+θ)\sin\theta\sin(60^\circ - \theta)\sin(60^\circ + \theta)

    =12sin⁡θ.2sin⁡(60∘−θ)sin⁡(60∘+θ)= \frac{1}{2}\sin\theta.2\sin(60^\circ - \theta)\sin(60^\circ + \theta)

    =12sin⁡θ[cos⁡2θ−cos⁡2π3]= \frac{1}{2}\sin\theta\left[\cos2\theta - \cos\frac{2\pi}{3}\right]

    =12sin⁡θcos⁡2θ+14sin⁡θ= \frac{1}{2}\sin\theta\cos2\theta + \frac{1}{4}\sin\theta

    =14(sin⁡3θ−sin⁡θ)+14sin⁡θ= \frac{1}{4}(\sin3\theta - \sin\theta) + \frac{1}{4}\sin\theta

    =14sin⁡3θ= \frac{1}{4}\sin3\theta

    Thus, we have the desired result.

  76. Let p=cos⁡αcos⁡βp = \cos\alpha\cos\beta

    =122cos⁡αcos⁡β=12[cos⁡(α+β)+cos⁡(α−β)]= \frac{1}{2}2\cos\alpha\cos\beta = \frac{1}{2}\left[\cos(\alpha + \beta) + \cos(\alpha - \beta)\right]

    =12cos⁡(α−β)[∵α+β=90∘∴cos⁡(α+β)=0]= \frac{1}{2}\cos(\alpha - \beta)[\because \alpha + \beta = 90^\circ \therefore \cos(\alpha + \beta) = 0]

    Now maximum value of cos⁡(α−β)\cos(\alpha - \beta) is 11 therefore maximum value of pp is 12\frac{1}{2}

  77. Since cos⁡α=12,\cos\alpha = \frac{1}{\sqrt{2}}, therefore α\alpha lies either in first quadrant or fourth quadrant. So sin⁡α=±12\sin\alpha = \pm\frac{1}{\sqrt{2}}

    We have to compute tan⁡α+β2cot⁡α−β2=5+26\tan\frac{\alpha + \beta}{2}\cot\frac{\alpha - \beta}{2} = 5 + 2\sqrt{6}

    =sin⁡α+β2cos⁡α−β2cos⁡α+β2sin⁡α−β2= \frac{\sin\frac{\alpha + \beta}{2}\cos\frac{\alpha - \beta}{2}}{\cos\frac{\alpha + \beta}{2}\sin\frac{\alpha - \beta}{2}}

    =sin⁡α+sin⁡βsin⁡α−sin⁡β= \frac{\sin\alpha + \sin\beta}{\sin\alpha - \sin\beta}

    Substituting the two pair of values, we get the desired answer.

  78. Let xcos⁡θ=ycos⁡(θ+2π3)=zcos⁡(θ+4π3)=kx\cos\theta = y\cos\left(\theta + \frac{2\pi}{3}\right) = z\cos\left(\theta + \frac{4\pi}{3}\right) = k

    Let p=kx+ky+kzp = \frac{k}{x} + \frac{k}{y} + \frac{k}{z}

    =cos⁡θ+cos⁡(θ+2π3)+cos⁡(θ+4π3)= \cos\theta + \cos\left(\theta + \frac{2\pi}{3}\right) + \cos\left(\theta + \frac{4\pi}{3}\right)

    =cos⁡θ+2cos⁡(θ+π)cos⁡π3=0[∵cos⁡π3=12 and cos⁡(θ+π)=−cos⁡θ]= \cos\theta + 2\cos(\theta + \pi)\cos\frac{\pi}{3} = 0[\because \cos\frac{\pi}{3} = \frac{1}{2}~\text{and}~\cos(\theta + \pi) = -\cos\theta]

    Thus, xy+yz+zx=0xy + yz + zx = 0

  79. Given, sin⁡θ=nsin⁡(θ+2α)\sin\theta = n\sin(\theta + 2\alpha)

    ⇒1n=sin⁡(θ+2α)sin⁡θ\Rightarrow \frac{1}{n} = \frac{\sin(\theta + 2\alpha)}{\sin\theta}

    Using componendo and dividendo

    ⇒1+n1−n=sin⁡(θ+2α)+sin⁡θsin⁡(θ+2α)−sin⁡θ\Rightarrow \frac{1 + n}{1 - n} = \frac{\sin(\theta + 2\alpha) + \sin\theta}{\sin(\theta + 2\alpha) - \sin\theta}

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

    =tan⁡(θ+α)cot⁡α= \tan(\theta + \alpha)\cot\alpha

  80. Given, sin⁡(θ+α)cos⁡(θ−α)=1−m1+m\frac{\sin(\theta + \alpha)}{\cos(\theta - \alpha)} = \frac{1 - m}{1 + m}

    Using componendo and dividendo

    ⇒sin⁡(θ+α)+cos⁡(θ−α)sin⁡(θ+α)−cos⁡(θ−α)=1−m+1+m1−m−1−m\Rightarrow \frac{\sin(\theta + \alpha) + \cos(\theta - \alpha)}{\sin(\theta + \alpha) - \cos(\theta - \alpha)} = \frac{1 - m + 1 + m}{1 - m - 1 - m}

    ⇒sin⁡(θ+α)+sin⁡(π2−(θ−α))sin⁡(θ+α)−sin⁡(π2−(θ−α))=2−2m=−1m\Rightarrow \frac{\sin(\theta + \alpha) + \sin\left(\frac{\pi}{2} - (\theta - \alpha)\right)}{\sin(\theta + \alpha) - \sin\left(\frac{\pi}{2} - (\theta - \alpha)\right)} = \frac{2}{-2m} = \frac{-1}{m}

    ⇒sin⁡(π4+α)cos⁡(θ−π4)cos⁡(π4+α)sin⁡(θ−π4)=−1m\Rightarrow \frac{\sin\left(\frac{\pi}{4} + \alpha\right)\cos\left(\theta - \frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4} + \alpha\right)\sin\left(\theta - \frac{\pi}{4}\right)} = -\frac{1}{m}

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

  81. Given, ysin⁡ϕ=xsin⁡(2θ+ϕ)y\sin\phi = x\sin(2\theta + \phi)

    sin⁡ϕsin⁡(2θ+ϕ)=xy\frac{\sin\phi}{\sin(2\theta + \phi)} = \frac{x}{y}

    By componendo and dividendo

    sin⁡ϕ+sin⁡(2θ+ϕ)sin⁡(2θ+ϕ)−sin⁡ϕ=x+yy−x\frac{\sin\phi + \sin(2\theta + \phi)}{\sin(2\theta + \phi) - \sin\phi} = \frac{x + y}{y - x}

    ⇒2sin⁡(θ+ϕ)cos⁡θ2cos⁡(θ+ϕ)sin⁡θ=x+yy−x\Rightarrow \frac{2\sin(\theta + \phi)\cos\theta}{2\cos(\theta + \phi)\sin\theta} = \frac{x + y}{y - x}

    ⇒cot⁡θcot⁡(θ+ϕ)=x+yy−x\Rightarrow \frac{\cot\theta}{\cot(\theta + \phi)} = \frac{x + y}{y - x}

    Hence, proven.

  82. Given, cos⁡(α+β)sin⁡(γ+δ)=cos⁡(α−beta)sin⁡(γ−δ)\cos(\alpha + \beta)\sin(\gamma + \delta) = \cos(\alpha - beta)\sin(\gamma - \delta)

    ⇒cos⁡(α+β)cos⁡(α−β)=sin⁡(γ−δ)sin⁡(γ+δ)\Rightarrow \frac{\cos(\alpha + \beta)}{\cos(\alpha - \beta)} = \frac{\sin(\gamma - \delta)}{\sin(\gamma + \delta)}

    By componendo and dividendo

    ⇒cos⁡(α+β)+cos⁡(α−β)cos⁡(α−β)−cos⁡(α+β)=sin⁡(γ−δ)+sin⁡(γ+δ)sin⁡(γ+δ)−sin⁡(γ−δ)\Rightarrow \frac{\cos(\alpha + \beta) + \cos(\alpha - \beta)}{\cos(\alpha - \beta) - \cos(\alpha + \beta)} = \frac{\sin(\gamma - \delta) + \sin(\gamma + \delta)}{\sin(\gamma + \delta) - \sin(\gamma - \delta)}

    ⇒2cos⁡αcos⁡β2sin⁡αsin⁡β=2sin⁡γcos⁡δ2cos⁡γsin⁡δ\Rightarrow \frac{2\cos\alpha\cos\beta}{2\sin\alpha\sin\beta} = \frac{2\sin\gamma\cos\delta}{2\cos\gamma\sin\delta}

    ⇒cot⁡αcot⁡β=cot⁡δcos⁡γ\Rightarrow \cot\alpha\cot\beta = \frac{\cot\delta}{\cos\gamma}

    Hence, proven.

83, 84 and 85 can be solved by using componendo and dividendo as well.