11. Transformation Formulae Solutions

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

   $= \frac{2\cos \frac{75^\circ + 15^\circ}{2}\sin \frac{75^\circ - 15^\circ}{2}}{2\cos \frac{75^\circ + 15^\circ}{2}\cos \frac{75^\circ - 15^\circ}{2}}$

   $= \frac{2\cos 45^\circ\sin30^\circ}{2\cos45^\circ\cos30^\circ} = \tan30^\circ = \frac{1}{\sqrt{3}}$

2. Given, $\frac{(\cos \theta - \cos 2\theta)(\sin 8\theta + \sin 2\theta)}{(\sin 5\theta - \sin\theta)(\cos 4\theta - \cos 6\theta)} = \frac{2\sin2\theta\sin\theta.2\sin5\theta\cos3\theta}{2\cos3\theta\sin2\theta.2\sin5\theta\sin\theta}$

   $= 1$

3. We have to prove, $\frac{\sin7\theta - \sin5\theta}{\cos7\theta + \cos5\theta} = \tan\theta$

   L.H.S. $= \frac{2\cos6\theta\sin\theta}{2\cos6\theta\cos\theta} = \tan\theta$

4. We have to prove, $\frac{\cos6\theta - \cos4\theta}{\sin6\theta + \sin4\theta} = -\tan\theta$

   L.H.S. $= \frac{2\sin5\theta\sin(-\theta)}{2\sin5\theta\cos\theta} = -\tan\theta$

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

   L.H.S. $= \frac{2\sin2A\cos(-A)}{2\cos2A\cos(-A)} = \tan 2A$

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

   L.H.S. $= \frac{2\cos4A\sin3A}{2\cos5A\sin3A} = \cos4A\sec5A$

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

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

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

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

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

9. We have to prove, $\frac{\sin A + \sin 2A}{\cos A - \cos 2A} = \cot \frac{A}{2}$

   L.H.S. $= \frac{2\sin\frac{3A}{2}\cos\frac{A}{2}}{2\sin\frac{3A}{2}\sin\frac{A}{2}}$

   $= \cot\frac{A}{2}$

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

    L.H.S. $= \frac{2\cos4A\sin A}{2\cos4A\cos A} = \tan A$

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

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

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

    L.H.S. $= \cos A\cos B - \sin A\sin B + \sin A\cos B - \cos A\sin B$

    $= (\sin A + \cos A)(\cos B - \sin B)$

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

    $= 2(\sin A\cos 45^\circ + \sin 45^\circ\sin A)(\cos 45^\circ\cos B - \sin 45^\circ\sin B)$

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

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

    L.H.S. $= \frac{-2\sin 2A\sin A}{2\cos 2A\sin A} + \frac{2\sin 3A\sin A}{2\cos 3A\sin A}$

    $= \frac{-\sin 2A}{\cos 2A} + \frac{\sin 3A}{\cos 3A}$

    $= \frac{\cos 2A\sin 3A - \sin 2A\cos 3A}{\cos 2A\cos 3A} = \frac{\sin(3A - 2A)}{\cos 2A\cos 3A}$

    $= \frac{\sin A}{\cos 3A\cos 3A}$

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

15. We have to prove, $\frac{\tan 5\theta + \tan 3\theta}{\tan 5\theta - \tan 3\theta} = 4\cos 2\theta\cos 4\theta$

    L.H.S. $= \frac{\frac{\sin5\theta}{\cos5\theta} + \frac{\sin4\theta}{\cos3\theta}}{\frac{\sin5\theta}{\cos5\theta} - \frac{\sin4\theta}{\cos3\theta}}$

    $= \frac{\sin5\theta\cos3\theta + \sin3\theta\cos5\theta}{\sin5\theta\cos3\theta - \sin3\theta\cos5\theta}$

    $= \frac{\sin8\theta}{\sin2\theta} = \frac{2\sin4\theta\cos\theta}{\sin2\theta}$

    $= \frac{4\sin2\theta\cos2\theta\cos4\theta}{\sin2\theta} = 4\cos2\theta\cos4\theta$

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

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

    L.H.S. $= \frac{2\cos5\theta\cos2\theta + 2\cos5\theta}{2\cos3\theta\cos2\theta + 2\cos3\theta}$

    $= \frac{\cos5\theta(\cos2\theta + 1)}{\cos3\theta(\cos2\theta + 1)}$

    $= \frac{\cos3\theta\cos2\theta - \sin3\theta\sin2\theta}{\cos3\theta}$

    $= \cos2\theta - \sin2\theta\tan3\theta$

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

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

    L.H.S. $= \frac{2\sin4A\cos3A + 2\sin4A\cos A}{2\cos4A\cos3A + 2\cos4A\cos A}$

    $= \frac{2\sin4A(\cos 3A + \cos A)}{2\cos4A(\cos 3A + \cos A)}$

    $= \tan 4A$

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

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

    L.H.S. $= \frac{2\sin\theta\cos\phi + 2\sin\theta}{2\cos\theta\cos\phi + 2\cos\theta}$

    $= \frac{2\sin\theta(\cos\phi + 1)}{2\cos\theta(\cos\phi + 1)}$

    $= \tan\theta$

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

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

    L.H.S. $= \frac{2\sin3A\cos2A + 2\sin3A}{2\sin5A\cos2A + 2\sin5A}$

    $= \frac{\sin3A(\cos 2A + 1)}{\sin5A(\cos 2A + 1)}$

    $= \frac{\sin 3A}{\sin 5A}$

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

    L.H.S. $= \frac{2\sin A\cos C + 2\sin A}{2\sin B\cos C + 2\sin B}$

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

    $= \frac{\sin A}{\sin B}$

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

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

    L.H.S. $= \frac{-2\cos7A\sin6A + 2\cos7A\sin 2A}{3\cos7A\cos6A - 2\cos7A\cos2A}$

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

    $= \frac{-2\cos 4A\sin 2A}{-2\sin 4A\sin 2A}$

    $= \cot 4A$

22. We have to prove that, $\frac{\sin A + \sin B}{\sin A - \sin B} = \tan \frac{A + B}{2}\cot \frac{A - B}{2}$

    L.H.S. $= \frac{2\sin\frac{A + B}{2}\cos\frac{A - B }{2}}{2\cos\frac{A + B}{2}\sin\frac{A - B}{2}}$

    $= \tan \frac{A + B}{2}\cot \frac{A - B}{2}$

23. We have to prove that, $\frac{\cos A + \cos B}{\cos B - \cos A} = \cot \frac{A + B}{2}\cot \frac{A - B}{2}$

    L.H.S. $= \frac{2\cos\frac{A + B}{2}\cos\frac{A - B}{2}}{2\sin\frac{A + B}{2}\sin\frac{A - B}{2}}$

    $= \cot\frac{A + B}{2}\cot\frac{A - B}{2}$

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

    L.H.S. $= \frac{2\sin\frac{A + B}{2}\cos\frac{A - B}{2}}{2\cos\frac{A + B}{2}\cos\frac{A - B}{2}}$

    $= \tan \frac{A + B}{2}$

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

    L.H.S. $= \frac{2\cos\frac{A + B}{2}\sin\frac{A - B}{3}}{2\sin\frac{A + B}{2}\sin\frac{A - B}{2}}$

    $= \cot \frac{A + B}{2}$

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

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

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

    $= \frac{2\cos B\cos C}{2\sin B\cos C} = \cot B$

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

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

    L.H.S. $= 2\cos9A\cos6A + 2\cos6A\cos A$

    $= 2\cos6A(\cos9A + \cos A)$

    $= 4\cos 4A \cos 5A \cos 6A$

28. We have to prove that, $\cos(-A + B + C) + \cos(A - B + C) + \cos(A + B - C) + \cos(A + B + C) = 4\cos A\cos B\cos C$

    Adding first two and last two, we get

    L.H.S. $= 2\cos C\cos (B - A) + 2\cos (A + B)\cos C$

    $= 2\cos C(\cos (B - A) + \cos (A + B))$

    $= 4\cos A\cos B\cos C$

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

    L.H.S. $= -2\cos 60^\circ \sin 10^\circ + \sin 10^\circ$

    $= -\sin10^\circ + \sin10^\circ = 0$

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

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

    $= 2\sin 30^\circ \cos 20^\circ + 2\sin 30^\circ \cos 10^\circ$

    $= 2\sin30^\circ(\cos 20^\circ + \cos 10^\circ)$

    $= \sin 70^\circ + \sin 80^\circ[\because \cos\theta = \sin(90^\circ - \theta)]$

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

    L.H.S. $= \sin\alpha + \sin 5\alpha + \sin2\alpha + \sin4\alpha$

    $= 2\sin3\alpha\cos2\alpha + 2\sin3\alpha\cos\alpha$

    $= 2\sin3\alpha(\cos2\alpha + \cos\alpha)$

    $= 4\cos \frac{\alpha}{2}\cos \frac{3\alpha}{2}\sin 3\alpha$

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

    $= 2\sin\left[\theta + n\phi\right]\sin\left[\frac{\phi}{2}\right]$

33. Given, $\sin\left[\theta + \left(n - \frac{3}{2}\right)\phi\right] + \sin\left[\theta + \left(n + \frac{3}{2}\right)\phi\right]$

    $= 2\sin\left[\theta + n\phi\right]\cos\left[\frac{\phi}{2}\right]$

34. Given, $2\sin5\theta\sin7\theta$

    Let the angles are $A$ and $B$ then $\cos C - \cos D = 2\sin \frac{C + D}{2}\cos\frac{D - C}{2}$

    Thus, comparing, we get

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

    $D = 12, C = 2$

    Thus, required expression is $\cos 2\theta - \cos12\theta$

35. Given, $2\cos7\theta\sin5\theta$

    $= \sin 12\theta + \sin 2\theta$

36. Given, $2\cos 11\theta\cos 3\theta$

    $= \cos 14\theta + \cos 8\theta$

37. Given, $2\sin54^\circ\sin66^\circ$

    $=\cos12^\circ - \cos120^\circ$

38. We have to prove that $\sin\frac{\theta}{2}\sin\frac{7\theta}{2} + \sin \frac{3\theta}{2}\sin\frac{11\theta}{2} =\sin 2\theta\sin 5\theta$

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

    $= \sin2\theta + \sin5\theta$

39. We have to prove that, $\cos 2\theta\cos \frac{\theta}{2} -\cos3\theta\cos\frac{9\theta}{2} = \sin5\theta\sin\frac{5\theta}{2}$

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

    $= \frac{1}{2}\left(2\sin 5\theta \sin \frac{5\theta}{2}\right)$

    $= \sin5\theta\sin\frac{5\theta}{2}$

40. We have to prove that, $\sin A\sin(A + 2B) - \sin B\sin(B + 2A) = \sin(A - B)\sin(A + B)$

    L.H.S. $= \frac{1}{2}(2\sin A\sin(A + 2B) - 2\sin B\sin(B + 2A))$

    $= \frac{1}{2}(\cos B - \cos(A + B) - \cos A - \cos (A + B))$

    $= \frac{1}{2}2\sin(A - B)\sin(A + B)$

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

    L.H.S. $= 2\sin2A\cos A\sin A - 2\sin2A\sin A\cos A = 0$

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

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

    $= \frac{\sin A}{\sin B}$

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

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

    $= \frac{\cos A - \cos 17A}{\sin 17A - \sin A}$

    $= \frac{2\sin 8A\sin 9A}{2\cos 9A\sin 8A} = \tan 9A$

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

    L.H.S. $= \frac{\cos 5A + \cos A - \cos 9A - \cos 5A + \cos 11A + \cos 9A}{\cos A - \cos 7A - \cos 3A + cos 7A + \cos 3A - \cos 11A}$

    $= \frac{\cos A + \cos 11A}{\cos A - \cos 11A}$

    $= \frac{\cos 6A\cos 5A}{\sin 6A\sin 5A} = \cot 6A\cot 5A$

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

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

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

    $= \cos 2A$

46. We have to prove that $\cos A\sin(B - C) + \cos B\sin(C - A) + \cos C\sin(A - B) = 0$

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

47. $\sin(45^\circ + A)\sin(45^\circ - A) = \frac{1}{2}\cos 2A$

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

48. We have to prove that, $\sin(\beta - \gamma)\cos(\alpha - \delta) + \sin(\gamma - \alpha)\cos(\beta - \delta) + \sin(\alpha - \beta)\cos(\gamma - \delta) = 0$

    L.H.S. $= \frac{1}{2}[\sin(\alpha + \beta - \gamma - \delta) + \sin(\beta + \delta - \gamma - \alpha) + \sin(\gamma -\alpha + \beta - \delta) + \sin(\gamma - \alpha - \beta + \delta) + \sin(\alpha - \beta + \gamma - \delta) - \sin(\alpha - \beta - \gamma + \delta)]$

    $= 0$

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

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

    $= \cos \frac{10\pi}{13} + \cos \frac{8\pi}{13} - \cos \frac{10\pi}{13} - \cos \frac{8\pi}{13} = 0$

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

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

    $= 2.\frac{1}{2}.\cos5^\circ - \cos 5^\circ = 0$

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

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

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

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

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

    $= \frac{2\sin 3A\cos 2A + 2\sin 3A\cos A}{2\cos3A\cos2A + 2\cos 3A\cos A}$

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

53. L.H.S $= \left(\frac{2\cos \frac{A + B}{2}\cos \frac{A - B}{2}}{2\cos\frac{A + B}{2}\sin\frac{A - B}{2}}\right)^n + \left(\frac{2\sin\frac{A + B}{2}\cos\frac{A - B}{2}}{2\sin\frac{A + B}{2}\sin\frac{B - A}{2}}\right)^n$

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

    $=\cot^n\frac{A - B}{2}[1 + (-1)^n]$ which is $0$ if $n$ is odd and $2\cos^n\frac{A - B}{2}$ if $n$ is even.

54. Given $\alpha, \beta, \gamma$ are in A.P. $\therefore 2\beta = \alpha + \gamma$

    R.H.S. $= \frac{\sin\alpha - \sin\gamma}{\cos\gamma - \cos\alpha} = \frac{2\cos \frac{\alpha + \gamma}{2}\sin\frac{\alpha - \gamma}{2}}{2\sin\frac{\gamma + \alpha}{2}\sin\frac{\alpha - \gamma}{2}}$

    $= \cot\frac{\alpha + \gamma}{2} = \cot\beta$

55. Given $\sin\theta + \sin\phi = \sqrt{3}(\cos\phi - \cos\theta)$

    $2\sin\frac{\theta + \phi}{2}\cos\frac{\theta - \phi}{2} = \sqrt{3}.2.\sin\frac{\theta + \phi}{2}\sin\frac{\theta - \phi}{2}$

    $= \sin\frac{\theta + \phi}{2}\left[\cos\frac{\theta - \phi}{2} - \sqrt{3}\sin\frac{\theta - \phi}{}\right] = 0$

    $\therefore \sin\frac{\theta + \phi}{2} = 0$ or $\tan\frac{\theta - \phi}{2} = \frac{1}{\sqrt{3}}$

    $\therefore \theta + \phi = 0^\circ$ or $\theta - \phi = 60^\circ$

    Now, $\sin3\theta + \sin3\phi = 2\sin\frac{3(\theta + \phi)}{2}\sin\frac{3(\theta - \phi)}{2} = 0$

    [$\because$ when $\theta + \phi = 0; \sin\frac{3(\theta + \phi)}{2} = 0$ and when $\theta - \phi = 60^\circ; \cos\frac{3(\theta - \phi)}{2} = 0$ ]

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

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

    $= 2\cos 45^\circ\cos20^\circ = 2.\frac{1}{\sqrt{2}}\cos20^\circ = \sqrt{2}\cos20^\circ$

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

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

    $= \cos 43^\circ + \cos77^\circ$

    $=2\cos 60^\circ \cos 17^\circ$

    $= \cos 17^\circ[\because 2\cos60^\circ = 2.\frac{1}{2} = 1$

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

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

    $= \frac{\cos10^\circ - \cos80^\circ}{\cos10^\circ + \cos80^\circ}$

    $= \frac{2\cos45^\circ\cos35^\circ}{2\cos45^\circ\cos35^\circ} = \tan 35^\circ$

59. We have to prove that, $\cos 80^\circ + \cos 40^\circ - cos 20^\circ = 0$

    L.H.S. $= 2\cos60^\circ\cos20^\circ - \cos20^\circ$

    $= \cos20^\circ - \cos20^\circ[\because 2\cos60^\circ = 2.\frac{1}{2} = 1]$

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

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

    $= 2\cos\frac{4\pi}{5}\cos\frac{3\pi}{5} + 2\cos\frac{4\pi}{5}\cos\frac{2\pi}{5}$

    $= 2\cos\frac{4\pi}{5}\left[\cos\frac{3\pi}{5} + \cos \left(\pi - \frac{3\pi}{5}\right)\right]$

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

61. We have to prove that $\cos\alpha + \cos\beta + \cos\gamma + \cos(\alpha + \beta + \gamma) = 4\cos\frac{\alpha + \beta}{2}\cos\frac{\beta + \gamma}{2}\cos \frac{\gamma + \alpha}{2}$

    L.H.S. $= \cos(\alpha + \beta + \gamma) + \cos\alpha + \cos\beta + \cos\gamma$

    $= 2\cos\left(\alpha + \frac{\beta + \gamma}{2}\right)\cos\frac{\beta + \gamma}{2} + 2\cos\frac{\beta+\gamma}{2}\cos\frac{\beta - \gamma}{2}$

    $= 2\cos\frac{\beta + \gamma}{2}\left[2\cos\left(\alpha +\frac{\beta + \gamma}{2}\right) + \cos\frac{\beta - \gamma}{2}\right]$

    $=4\cos\frac{\alpha + \beta}{2}\cos\frac{\beta + \gamma}{2}\cos \frac{\gamma + \alpha}{2}$

62. Given, $\sin\alpha - \sin\beta = \frac{1}{3}$ and cosbeta - cosalpha = frac{1}{2}

    Dividing we get, $\frac{\sin\alpha - \sin\beta}{\cos\beta - \cos\alpha} = \frac{2}{3}$

    $\Rightarrow \frac{2\cos\frac{\alpha + \beta}{2}\sin\frac{\alpha - \beta}{2}}{2\sin\frac{\alpha + \beta}{2}\sin\frac{\alpha - \beta}{2}} = \frac{2}{3}$

    $\Rightarrow \cot\frac{\alpha + \beta}{2} = \frac{2}{3}$

63. Given, $\cosec A + sec A = \cosec B + \sec B$

    $\sec A - \sec B = \cosec B - \cosec A$

    $\Rightarrow \frac{\cos B - \cos A}{\cos A\cos B} = \frac{\sin A - \sin B}{\sin A\sin B}$

    $\Rightarrow \frac{2\sin\frac{A + B}{2}\sin \frac{A - B}{2}}{\cos A\cos B} = \frac{2\cos\frac{A + B}{2}\sin\frac{A - B}{2}}{\sin A\sin B}$

    $\Rightarrow \tan A\tan B = \cot \frac{A + B}{2}$

64. Given, $\sec(\theta + \alpha) + \sec(\theta - \alpha) = 2\sec\theta$

    $\Rightarrow \frac{1}{\cos(\theta + \alpha)} + \frac{1}{\cos(\theta - \alpha)} = \frac{2}{\cos\theta}$

    $\cos\theta[\cos(\theta - \alpha) + \cos(\theta + \alpha)] = 2\cos(\theta + \alpha)\cos(\theta - \alpha)$

    $\cos\theta.2\cos\theta\cos\alpha = \cos2\theta + \cos2\alpha$

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

    Thus, the above equation becomes

    $2\cos^2\theta\cos\alpha = 2\cos^2\theta - 1 + 2\cos^2\alpha - 1$

    $2\cos^2\theta(\cos\alpha - 1) = 2(\cos^2\alpha - 1)$

    $\Rightarrow \cos^2\theta = 1 + \cos\alpha$

65. We have to prove that $\sin50^\circ\cos85^\circ = \frac{1 - \sqrt{2}\sin 35^\circ}{2\sqrt{2}}$

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

    $= \frac{1}{2}[\sin135^\circ - \sin35^\circ]$

    $= \frac{1}{2}\left[\frac{1}{\sqrt{2}} - \sin35^\circ\right]$

    $= \frac{1 - \sqrt{2}\sin 35^\circ}{2\sqrt{2}}$

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

    L.H.S. $= \frac{1}{2}(2\sin80^\circ\sin40^\circ)\sin20^\circ$

    $= \frac{1}{2}[\cos(80^\circ - 40^\circ) - \cos(80^\circ + 40^\circ)]\sin20^\circ$

    $= \frac{1}{2}(\cos40^\circ - \cos120^\circ)\sin 206\circ$

    $= \frac{1}{4}(2\cos40^\circ\sin20^\circ - 2\cos120^\circ\sin20^\circ)$

    $= \frac{1}{4}[\sin(40^\circ + 20^\circ) - \sin(40^\circ - 20^\circ) - 2.-\frac{1}{2}\sin20^\circ]$

    $= \frac{1}{4}[\sin60^\circ - \sin20^\circ + \sin20^\circ] = \frac{1}{4}\sin60^\circ = \frac{\sqrt{3}}{8}$

67. We have to prove that, $\sin A\sin(60^\circ - A)\sin(60^\circ + A) = \frac{1}{4}\sin 3A$

    L.H.S. $= \frac{1}{2}\sin A[2\sin(60^\circ - A)\sin(60^\circ + A)]$

    $= \frac{1}{2}\sin A[\cos(60^\circ + A - 60^\circ + A) - \cos(60^\circ + A + 60^\circ - A)]$

    $= \frac{1}{2}\sin A(\cos 2A - \cos 120^\circ)$

    $= \frac{1}{4}(2\sin A\cos 2A - 2\cos 120^\circ \sin A)$

    $\frac{1}{4}[\sin (2A + A) - \sin(2A - A) - 2.-\frac{1}{2}\sin A]$

    $= \frac{1}{4}(\sin 3A - \sin A + \sin A) = \frac{1}{4}\sin 3A$

68. Let $p = \sin\alpha\sin\beta$

    $= \frac{1}{2}2\sin\alpha\sin\beta$