11. Transformation Formulae SolutionsΒΆ

  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 \(p = \sin\alpha\sin\beta\)

    \(= \frac{1}{2}2\sin\alpha\sin\beta\)

    \(= \frac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha + beta)]\)

    \(= \frac{1}{2}[\cos(\alpha - \beta) - \cos90^\circ][\because \alpha + \beta = 90^\circ~\text{(given)}]\)

    \(= \frac{1}{2}\cos(\alpha - \beta)\)

    Maximum value of \(\cos(\alpha - \beta)\) is \(1,\) hence maximum value of \(p\) is \(\frac{1}{2}\)

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

    L.H.S. \(= \sin 25^\circ\cos 115^\circ\)

    \(= \frac{1}{2}2\sin25^\circ\cos115^\circ = \frac{1}{2}[\sin140^\circ - \sin90^\circ]\)

    \(= \frac{1}{2}[\cos50^\circ - 1] = \frac{1}{2}[\sin40^\circ - 1]\)

  70. We have to prove that, \(\sin 20^\circ \sin 40^\circ\sin 60^\circ \sin80^\circ = \frac{3}{16}\)

    L.H.S. \(= \frac{1}{2}[(2\sin 20^\circ\sin80^\circ)(\sin40^\circ\sin60^\circ)]\)

    \(= \frac{1}{2}(\cos60^\circ - \cos100^\circ)\frac{\sqrt{3}}{2}\sin40^\circ\)

    \(= \frac{\sqrt{3}}{4}[\cos60^\circ\sin40^\circ + \sin10^\circ\sin40^\circ]\)

    \(= \frac{\sqrt{3}}{8}[2\cos60^\circ\sin40^\circ + 2\cos80^\circ\sin40^\circ]\)

    \(= \frac{\sqrt{3}}{8}[\sin100^\circ - \sin 20^\circ + \sin 120^\circ - \sin 40^\circ]\)

    \(= \frac{\sqrt{3}}{8}[\cos10^\circ - (\sin20^\circ + \sin40^\circ) + \cos30^\circ]\)

    \(= \frac{\sqrt{3}}{8}[\cos10^\circ - 2\sin30^\circ\cos10^\circ + \frac{\sqrt{3}}{2}]\)

    \(= \frac{3}{8}[\cos10^\circ - 2.\frac{1}{2}.\cos10^\circ + \frac{3}{2}] = \frac{3}{16}\)

  71. We have to prove that, \(\cos 20^\circ\cos40^\circ\cos80^\circ = \frac{1}{8}\)

    L.H.S. \(= \frac{1}{2\cos20^\circ\cos80^\circ}\cos40^\circ\)

    \(= \frac{1}{2}[\cos100^\circ + \cos60^\circ]\cos40^\circ\)

    \(= \frac{1}{2}[-\sin10^\circ\cos40^\circ + \frac{1}{2}\cos40^\circ]\)

    \(= \frac{1}{4}[-2\cos40^\circ\sin10^\circ + \cos40^\circ]\)

    \(= \frac{1}{4}[-\sin50^\circ + sin30^\circ + \cos40^\circ]\)

    \(= \frac{1}{8}[\because \sin50^\circ = \cos40^\circ~\text{and}~\sin30^\circ = \frac{1}{2}]\)

  72. We have to prove that, \(\tan20^\circ\tan40^\circ\tan60^\circ\tan80^\circ = 3\)

    Using results of 70 and 71 it can be solved.

  73. We have to prove that, \(\cos10^\circ\cos30^\circ\cos50^\circ\cos70^\circ = \frac{3}{16}\)

    L.H.S. \(= \frac{1}{2}(2\cos10^\circ\cos70^\circ)(\frac{\sqrt{3}}{2}\cos50^\circ)\)

    \(= \frac{\sqrt{3}}{4}[\cos80^\circ + \cos60^\circ]\cos50^\circ\)

    \(= \frac{\sqrt{3}}{4}[\cos80^\circ\cos50^\circ + \frac{1}{2}\cos50^\circ]\)

    \(= \frac{\sqrt{3}}{8}[2\cos80^\circ\cos50^\circ + \cos50^\circ]\)

    \(= \frac{\sqrt{3}}{8}[\cos 130^\circ + \cos30^\circ + \cos50^\circ]\)

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

    \(= \frac{3}{16}\)

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

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

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

    \(= -\cos\theta + 2\cos\theta\cos2\theta = -\cos\theta + \cos3\theta + \cos\theta = \cos3\theta\)

  75. We have to prove that \(\tan\theta\tan(60^\circ - \theta)\tan(60^\circ + \theta) = \tan3\theta\)

    We have just proven, \(4\cos\theta\cos\left(\frac{\pi}{3} + \theta\right)\cos\left(\frac{\pi}{3} - \theta\right) = \cos3\theta\)

    Let us evaluate \(\sin\theta\sin(60^\circ - \theta)\sin(60^\circ + \theta)\)

    \(= \frac{1}{2}\sin\theta.2\sin(60^\circ - \theta)\sin(60^\circ + \theta)\)

    \(= \frac{1}{2}\sin\theta\left[\cos2\theta - \cos\frac{2\pi}{3}\right]\)

    \(= \frac{1}{2}\sin\theta\cos2\theta + \frac{1}{4}\sin\theta\)

    \(= \frac{1}{4}(\sin3\theta - \sin\theta) + \frac{1}{4}\sin\theta\)

    \(= \frac{1}{4}\sin3\theta\)

    Thus, we have the desired result.

  76. Let \(p = \cos\alpha\cos\beta\)

    \(= \frac{1}{2}2\cos\alpha\cos\beta = \frac{1}{2}\left[\cos(\alpha + \beta) + \cos(\alpha - \beta)\right]\)

    \(= \frac{1}{2}\cos(\alpha - \beta)[\because \alpha + \beta = 90^\circ \therefore \cos(\alpha + \beta) = 0]\)

    Now maximum value of \(\cos(\alpha - \beta)\) is \(1\) therefore maximum value of \(p\) is \(\frac{1}{2}\)

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

    We have to compute \(\tan\frac{\alpha + \beta}{2}\cot\frac{\alpha - \beta}{2} = 5 + 2\sqrt{6}\)

    \(= \frac{\sin\frac{\alpha + \beta}{2}\cos\frac{\alpha - \beta}{2}}{\cos\frac{\alpha + \beta}{2}\sin\frac{\alpha - \beta}{2}}\)

    \(= \frac{\sin\alpha + \sin\beta}{\sin\alpha - \sin\beta}\)

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

  78. Let \(x\cos\theta = y\cos\left(\theta + \frac{2\pi}{3}\right) = z\cos\left(\theta + \frac{4\pi}{3}\right) = k\)

    Let \(p = \frac{k}{x} + \frac{k}{y} + \frac{k}{z}\)

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

    \(= \cos\theta + 2\cos(\theta + \pi)\cos\frac{\pi}{3} = 0[\because \cos\frac{\pi}{3} = \frac{1}{2}~\text{and}~\cos(\theta + \pi) = -\cos\theta]\)

    Thus, \(xy + yz + zx = 0\)

  79. Given, \(\sin\theta = n\sin(\theta + 2\alpha)\)

    \(\Rightarrow \frac{1}{n} = \frac{\sin(\theta + 2\alpha)}{\sin\theta}\)

    Using componendo and dividendo

    \(\Rightarrow \frac{1 + n}{1 - n} = \frac{\sin(\theta + 2\alpha) + \sin\theta}{\sin(\theta + 2\alpha) - \sin\theta}\)

    \(= \frac{\sin(\theta + \alpha)\cos\alpha}{\cos(\theta + \alpha)\sin\alpha}\)

    \(= \tan(\theta + \alpha)\cot\alpha\)

  80. Given, \(\frac{\sin(\theta + \alpha)}{\cos(\theta - \alpha)} = \frac{1 - m}{1 + m}\)

    Using componendo and dividendo

    \(\Rightarrow \frac{\sin(\theta + \alpha) + \cos(\theta - \alpha)}{\sin(\theta + \alpha) - \cos(\theta - \alpha)} = \frac{1 - m + 1 + m}{1 - m - 1 - m}\)

    \(\Rightarrow \frac{\sin(\theta + \alpha) + \sin\left(\frac{\pi}{2} - (\theta - \alpha)\right)}{\sin(\theta + \alpha) - \sin\left(\frac{\pi}{2} - (\theta - \alpha)\right)} = \frac{2}{-2m} = \frac{-1}{m}\)

    \(\Rightarrow \frac{\sin\left(\frac{\pi}{4} + \alpha\right)\cos\left(\theta - \frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4} + \alpha\right)\sin\left(\theta - \frac{\pi}{4}\right)} = -\frac{1}{m}\)

    \(\Rightarrow m = \tan\left(\frac{\pi}{4} - \theta\right)\tan\left(\frac{\pi}{4} - \alpha\right)\)

  81. Given, \(y\sin\phi = x\sin(2\theta + \phi)\)

    \(\frac{\sin\phi}{\sin(2\theta + \phi)} = \frac{x}{y}\)

    By componendo and dividendo

    \(\frac{\sin\phi + \sin(2\theta + \phi)}{\sin(2\theta + \phi) - \sin\phi} = \frac{x + y}{y - x}\)

    \(\Rightarrow \frac{2\sin(\theta + \phi)\cos\theta}{2\cos(\theta + \phi)\sin\theta} = \frac{x + y}{y - x}\)

    \(\Rightarrow \frac{\cot\theta}{\cot(\theta + \phi)} = \frac{x + y}{y - x}\)

    Hence, proven.

  82. Given, \(\cos(\alpha + \beta)\sin(\gamma + \delta) = \cos(\alpha - beta)\sin(\gamma - \delta)\)

    \(\Rightarrow \frac{\cos(\alpha + \beta)}{\cos(\alpha - \beta)} = \frac{\sin(\gamma - \delta)}{\sin(\gamma + \delta)}\)

    By componendo and dividendo

    \(\Rightarrow \frac{\cos(\alpha + \beta) + \cos(\alpha - \beta)}{\cos(\alpha - \beta) - \cos(\alpha + \beta)} = \frac{\sin(\gamma - \delta) + \sin(\gamma + \delta)}{\sin(\gamma + \delta) - \sin(\gamma - \delta)}\)

    \(\Rightarrow \frac{2\cos\alpha\cos\beta}{2\sin\alpha\sin\beta} = \frac{2\sin\gamma\cos\delta}{2\cos\gamma\sin\delta}\)

    \(\Rightarrow \cot\alpha\cot\beta = \frac{\cot\delta}{\cos\gamma}\)

    Hence, proven.

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