15. Trigonometrical Identities SolutionsΒΆ

  1. \(\because A + B + C = \pi \therefore A + B = \pi - C\)

    \(\Rightarrow \cos(A + B) = \cos(\pi - C) = \cos C \Rightarrow \sin A\sin B - \cos C = \cos A\cos B\)

    \(\Rightarrow (\sin A\sin B - \cos C)^2 = \cos^2A\cos^2B\)

    \(\Rightarrow \sin^2A\sin^2B + \cos^2C - 2\sin A\sin B\cos C = (1 - \sin^2A)(1 - \sin^2B)\)

    \(\Rightarrow \sin^A + \sin^2B + \cos^2C - 1 = 2\sin A\sin B\cos C\)

    \(\Rightarrow \sin^2A + \sin^2B - \cos^2C = 2\sin A\sin B\cos C\)

  2. \(A + B + C = 180^\circ \Rightarrow \frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^\circ \Rightarrow \frac{A}{2} + \frac{B}{2} = 90^\circ - \frac{C}{2}\)

    \(\Rightarrow \cos\left(\frac{A}{2} + \frac{B}{2}\right) = \cos\left(90^\circ - \frac{C}{2}\right)\)

    \(\Rightarrow \cos\frac{A}{2}\cos\frac{B}{2} - \sin\frac{A}{2}\sin\frac{B}{2} = \sin\frac{C}{2}\)

    \(\Rightarrow \sin\frac{C}{2} + \sin\frac{A}{2}\sin\frac{B}{2} = \cos\frac{A}{2}\cos\frac{B}{2}\)

    \(\Rightarrow \left(\sin\frac{C}{2} + \sin\frac{A}{2}\sin\frac{B}{2}\right)^2 = \cos^2\frac{A}{2}\cos^2\frac{B}{2}\)

    \(\Rightarrow \sin^2\frac{C}{2} + \sin^2\frac{A}{2}\sin^2\frac{B}{2} + 2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} = \left(1 - \cos^2\frac{A}{2}\right)\left(1 - \cos^2\frac{B}{2}\right)\)

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

  3. Let \(A + B = C \Rightarrow \cos(A + B) = \cos C\)

    \(\Rightarrow \sin A\sin B + \cos C = \cos A\cos B \Rightarrow (\sin A\sin B + \cos C)^2 = \cos^2A\cos^2B\)

    \(\Rightarrow \sin^2A\sin^2B + \cos^2C + 2\sin A\sin B\cos C = (1 - \sin^2A)(1 - \sin^2B)\)

    \(\Rightarrow \sin^2A + \sin^2B + 2\sin A\sin B\cos C = \sin^2C\)

    \(\Rightarrow \sin^2A + \sin^2B + 2\sin A\sin B\cos(A + B) = \sin^2(A + B)\)

  4. Given, \(A + B + C = 180^\circ \Rightarrow A + B = 180^\circ - C \Rightarrow \cos(A + B) = -\cos C\)

    \(\Rightarrow \cos A\cos B + \cos C = \sin A\sin B \Rightarrow (\cos A\cos B + \cos C)^2 = \sin^2A\sin^2B\)

    \(\Rightarrow \cos^2A\cos^2B + \cos^2C + 2\cos A\cos B\cos C = (1 - \cos^2A)(1 - \cos^2B)\)

    \(\Rightarrow \cos^2A + \cos^2B + \cos^2C + 2\cos A\cos B\cos C = 1\)

  5. We have just proved that \(\cos^2A + \cos^2B + \cos^2C + 2\cos A\cos B\cos C = 1\)

    \(\Rightarrow 3 - \sin^2A - \sin^2B - \sin^2C + 2\cos A\cos B\cos C = 1\)

    \(\Rightarrow \sin^2A + \sin^2B + \sin^2C = 2(1 + \cos A\cos B\cos C)\)

  6. Given, \(A + B + C = 180^\circ \Rightarrow A + B = 180^\circ - C \Rightarrow \cos(A + B) = -\cos C\)

    \(\Rightarrow \cos A\cos B = \sin A\sin B - \cos C \Rightarrow \cos^2A\cos^2B = \sin^2A\sin^2B + \cos^2C - 2\sin A\sin B\cos C\)

    \(\Rightarrow \cos^2A\cos^2B = (1 - \cos^2A)(1 - \cos^2B) + \cos^2C - 2\sin A\sin B\cos C\)

    \(\Rightarrow \cos^2A + \cos^2B - \cos^2C = 1 - 2\sin A\sin B\cos C\)

  7. Given, \(A + B + C = 180^\circ \Rightarrow \frac{A + B + C}{2} = 90^\circ\)

    \(\Rightarrow \cos\left(\frac{A}{2} + \frac{B}{2}\right) = \sin\frac{C}{2}\)

    \(\Rightarrow \cos\frac{A}{2}\cos\frac{B}{2} - \sin\frac{C}{2} = \sin\frac{A}{2}\sin\frac{B}{2}\)

    \(\Rightarrow \cos^2\frac{A}{2}\cos^2\frac{B}{2} + \sin^2\frac{C}{2} - 2\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2} = \left(1 - \cos^2\frac{A}{2}\right)\left(1 - \cos^2\frac{B}{2}\right)\)

    \(\Rightarrow \cos^2\frac{A}{2} + \cos^2\frac{B}{2} - \cos^2\frac{C}{2} = 2\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2}\)

  8. Given, \(A + B + C = 180^\circ \Rightarrow \frac{A + B + C}{2} = 90^\circ\)

    \(\Rightarrow \cos\left(\frac{A}{2} + \frac{B}{2}\right) = \sin\frac{C}{2}\)

    \(\Rightarrow \cos\frac{A}{2}\cos\frac{B}{2} = \sin\frac{A}{2}\sin\frac{B}{2} + \sin\frac{C}{2}\)

    \(\Rightarrow \cos^2\frac{A}{2}\cos^2\frac{B}{2} = \sin^2\frac{A}{2}\sin^2\frac{B}{2} + \sin^2\frac{C}{2} + 2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\)

    \(\Rightarrow \cos^2\frac{A}{2}\cos^2\frac{B}{2} = \left(1 - \cos^2\frac{A}{2}\right)\left(1 - \cos^2\frac{B}{2}\right) + \sin^2\frac{C}{2} + 2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\)

    \(\cos^2\frac{A}{2} + \cos^2\frac{B}{2} + \cos^2\frac{C}{2} = 2 + 2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\)

  9. Given, \(A + B + C = \frac{\pi}{2} \Rightarrow A + B = \frac{\pi}{2} - C \Rightarrow \cos(A + B) = \sin C\)

    \(\Rightarrow \cos A\cos B = \sin A\sin B + \sin C\)

    \(\Rightarrow \cos^2A\cos^2B = \sin^2A\sin^2B + \sin^2C + 2\sin A\sin B\sin C\)

    \(\Rightarrow (1 - \sin^2A)(1 - \sin^2B) = \sin^2A\sin^2B + \sin^2C + 2\sin A\sin B\sin C\)

    \(\Rightarrow \sin^2A + \sin^2B + \sin^2C = 1 - 2\sin A\sin B\sin C\)

  10. We have just proven that \(\sin^2A + \sin^2B + \sin^2C = 1 - 2\sin A\sin B\sin C\) in previous problem.

    \(\Rightarrow 1 - \cos^2A + 1 - \cos^2B + 1 - \cos^2C = 1 - 2\sin A\sin B\sin C\)

    \(\Rightarrow \cos^2A + \cos^2B + \cos^2C = 2 + 2\sin A\sin B\sin C\)

  11. Givem \(A + B + C = 2\pi \Rightarrow A + B = 2\pi - C \Rightarrow \cos(A + B) = \cos C\)

    \(\Rightarrow \cos A\cos B - \cos C = \sin A\sin B\)

    \(\Rightarrow \cos^2A\cos^2B + \cos^2C - 2\cos A\cos B\cos C = \sin^2A\sin^2B = (1 - \cos^2A)(1 - \cos^2B)\)

    \(\Rightarrow \cos^2A + \cos^2B + \cos^2C - 2\cos A\cos B\cos C = 1\)

  12. Given \(A + B = C \Rightarrow \cos(A + B) = \cos C\)

    \(\Rightarrow \cos A\cos B - \cos C = \sin A\sin B\)

    \(\Rightarrow \cos^2A\cos^2B + \cos^2C - 2\cos A\cos B\cos C = \sin^2A\sin^2B\)

    \(\Rightarrow \cos^2A\cos^2B + \cos^2C - 2\cos A\cos B\cos C = (1 - \cos^2A)(1 - \cos^2B)\)

    \(\Rightarrow \cos^2A + \cos^2B + \cos^2C - 2\cos A\cos B\cos C = 1\)

  13. Given \(A + B = \frac{\pi}{3} \Rightarrow \cos(A + B) = \cos\frac{\pi}{3} = \frac{1}{2}\)

    \(\Rightarrow \cos A\cos B - \frac{1}{2} = \sin A\sin B\)

    \(\Rightarrow \cos^2A\cos^2B - \cos A\cos B + \frac{1}{4} = \sin^2A\sin^2B = (1 - \cos^2A)(1 - \cos^2B)\)

    \(\Rightarrow \cos^2A + \cos^2B - \cos A\cos B = \frac{3}{4}\)

  14. From problem 12 we have \(A + B = C\) and \(\cos^2A + \cos^2B + \cos^2C - 2\cos A\cos B\cos C = 1\)

    Substituting \(C = A + B\) we get \(\cos^2A + \cos^2B + \cos^2(A + B) - 2\cos A\cos B\cos(A + B) = 1\)

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

  15. Given \(A + B + C = \pi\) and \(A + B = 2C \Rightarrow C = \frac{\pi}{3} \Rightarrow A + B = \pi - \frac{pi}{3}\)

    \(\cos(A + B) = -\cos\frac{\pi}{3}\Rightarrow \cos A\cos B = \sin A\sin B - \frac{1}{2}\)

    \(\Rightarrow \cos^2A\cos^2B = \sin^2A\sin^2B - \sin A\sin B + \frac{1}{4}\)

    \(\Rightarrow (1 - \sin^2A)(1 - \sin^2B) = \sin^2A\sin^2B - \sin A\sin B + \frac{1}{4}\)

    \(\Rightarrow 4(\sin^2A + \sin^2B - \sin A\sin B) = 3\)

  16. Given \(A + B + C = 2\pi \Rightarrow \cos(B + C) = \cos(2\pi - A) = \cos A\)

    \(\Rightarrow \cos B\cos C - \cos A = \sin B\sin C\)

    \(\Rightarrow \cos^B\cos^2C + \cos^2A - 2\cos A\cos B\cos C = \sin^2B\sin^2C = (1 - \cos^2B)(1 - \cos^2C)\)

    \(\Rightarrow \cos^2B + \cos^2C - \sin^2A - 2\cos A\cos B\cos C = 0\)

  17. Given \(A + B + C = 0 \Rightarrow \cos(A + B) = \cos C\)

    \(\Rightarrow \cos A\cos B - \cos C = \sin A\sin B\)

    \(\Rightarrow \cos^2A\cos^2B + \cos^2C - 2\cos A\cos B\cos C = \sin^2A\sin^2B = (1 - \cos^2A)(1 - \cos^2B)\)

    \(\Rightarrow \cos^2A + \cos^2B + \cos^2C = 1 + 2\cos A\cos B\cos C\)

  18. Putting \(A = B - C, B = C - A\) and \(C = A - B\) in 17 we can obtain the desired result.

  19. Given \(A + B + C = \pi,\) we have to prove that \(\sin A\cos B\cos C + \sin B\cos C\cos A + \sin C\cos A\cos B= \sin A\sin B\sin C\)

    Dividing both sides by \(\sin A\sin B\sin C,\) we get

    \(\cot B\cot C + \cot C\cot A + \cot A\cot B = 1\)

    \(A + B = \pi - C\Rightarrow \cot(A + B) = -\cot C\)

    \(\Rightarrow \frac{\cot A\cot B - 1}{\cot A + \cot B} = -\cot C\)

    \(\Rightarrow \cot B\cot C + \cot C\cot A + \cot A\cot B = 1\)

  20. Given, \(A + B + C = \pi \Rightarrow A + B = \pi - C\)

    \(\Rightarrow \tan(A + B) = \tan(\pi - C) = -\tan C\)

    \(\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan B} = -\tan C\)

    \(\Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C\)

  21. Given \(A + B + C = \pi \Rightarrow \frac{A + B}{2} = \frac{\pi - C}{2}\)

    \(\Rightarrow \tan\frac{A + B}{2} = \tan\frac{\pi - C}{2}\)

    \(\Rightarrow \frac{\tan\frac{A}{2} + \tan\frac{B}{2}}{1 - \tan\frac{A}{2}\tan\frac{B}{2}} = \cot\frac{C}{2} = \frac{1}{\tan\frac{C}{2}}\)

    \(\Rightarrow \tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1\)

  22. Let \(B + C - A = \alpha, C + A - B = \beta, A + B - C = \gamma\)

    \(\alpha + \beta + \gamma = A + B + C = \pi\)

    We have just proven that if \(A + B + C = \pi\) then \(\Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C\)

    Thus, substituting we get, \(\Rightarrow \tan\alpha + \tan\beta + \tan\gamma = \tan\alpha\tan\beta\tan\gamma\)

    \(\Rightarrow \tan(B + C - A) + \tan(C + A - B) + \tan(A + B - C) = \tan(B + C - A)\tan(C + A - B)\tan(A + B - C)\)

  23. Given \(A + B + C = \pi\Rightarrow A + B = \pi - C \Rightarrow \cot(A + B) = \cot(\pi - C)\)

    \(\Rightarrow \frac{\cot A\cot B - 1}{\cot A + \cot B} = -\cot C\)

    \(\Rightarrow \cot B\cot C + \cot C\cot A + \cot A\cot B = 1\)

  24. From previosu problem if \(A + B + C = \pi\) then \(\Rightarrow \cot B\cot C + \cot C\cot A + \cot A\cot B = 1\)

    Given \(\cot A + \cot B + \cot C = \sqrt{3}\)

    \(\Rightarrow \cot^2A + \cot^2B + \cot^2C + 2(\cot A\cot B + \cot B\cot C + \cot C\cot A) = 3\)

    \(\cot^2A + \cot^2B + \cot^2C = 1\)

    \(2\cot^2A + 2\cot^2B + 2\cot^2C - 2 = 0\)

    \(2\cot^2A + 2\cot^2B + 2\cot^2C - 2(\cot A\cot B + \cot B\cot C + \cot C\cot A) = 0\)

    \((\cot A - \cot B)^2 + (\cot B - \cot C)^2 + (\cot C - \cot A)^2 = 0\)

    This is possible only if \(\cot A - \cot B = 0\) i.e. \(\cot A = \cot B,\) \(\cot B - \cot C = 0\) i.e. \(\cot B = \cot C\) and \(\cot C - \cot A = 0\) i.e. \(\cot C = \cot A\)

    \(\therefore \cot A = \cot B = \cot C \Rightarrow A = B = C\)

  25. \(\because A + B + C + D = 2\pi \Rightarrow A + B = 2\pi - C - D\)

    \(\Rightarrow \tan(A + B) = -\tan(C + D)\)

    \(\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan B} = -\frac{\tan C + \tan D}{1 - \tan C\tan D}\)

    \(\Rightarrow (\tan A + \tan B)(1 - \tan C\tan D) = -(1 - \tan A\tan B)(\tan C + \tan D)\)

    \(\Rightarrow \tan A + \tan B + \tan C + \tan D = \tan A\tan B\tan C + \tan A\tan C\tan D + \tan A\tan B\tan D + \tan B\tan C\tan D\)

    Dividing both sides by \(\tan A\tan B\tan C\tan D,\) we get

    \(\frac{\tan A + \tan B + \tan C + \tan D}{\tan A\tan B\tan C\tan D} = \frac{1}{\tan A} + \frac{1}{\tan B} + \frac{1}{\tan C} + \frac{1}{\tan D}\)

    \(\Rightarrow \frac{\tan A + \tan B + \tan C + \tan D}{\cot A + \cot B + \cot C + \cot D} = \tan A\tan B\tan C\tan D\)

  26. Given \(A + B + C = \frac{\pi}{2}\Rightarrow A + B = \frac{\pi}{2} - C\)

    \(\Rightarrow \cot(A + B) = \cot\left(\frac{\pi}{2} - C\right)\)

    \(\Rightarrow \frac{\cot A\cot B - 1}{\cot A + \cot B} = \tan C = \frac{1}{\cot C}\)

    \(\Rightarrow \cot A + \cot B + \cot C = \cot A\cot B\cot C\)

  27. We have just proven in 26 that \(\Rightarrow \cot A + \cot B + \cot C = \cot A\cot B\cot C\)

    Dividing both sides by \(\cot A\cot B\cot C,\) we get

    \(\tan A\tan B + \tan B\tan C + \tan C\tan A = 1\)

  28. Given \(A + B + C = \pi \Rightarrow 3(A + B + C) = 3\pi \Rightarrow 3A + 3B = 3\pi - 3C\)

    \(\Rightarrow \tan(3A + 3B) = \tan(3\pi - 3C) = -\tan3C\)

    \(\Rightarrow \frac{\tan 3A + \tan 3B}{1 - \tan3A\tan3B} = -\tan3C\)

    \(\Rightarrow \tan 3A + \tan 3B + \tan 3C = \tan 3A\tan 3B\tan 3C\)

  29. Given \(A + B + C = \pi \Rightarrow \frac{A + B}{2} = \frac{\pi - C}{2}\)

    \(\Rightarrow \cot\frac{A + B}{2} = \cot\frac{\pi - C}{2}\)

    \(\Rightarrow \frac{\cot\frac{A}{2}\cot\frac{B}{2} - 1}{\cot\frac{A}{2} + \cot\frac{B}{2}} = \tan\frac{C}{2} = \frac{1}{\cot\frac{C}{2}}\)

    \(\Rightarrow \cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}\)

  30. We have to prove that \(\frac{\cot A + \cot B}{\tan A + \tan B} + \frac{\cot B + \cot C}{\tan B + \tan C} + \frac{\cot C + \cot A}{\tan C + \tan A} = 1\)

    Putting \(\tan A = \frac{1}{\cot A}, \tan B = \frac{1}{\cot B}, \tan C = \frac{1}{\cot C},\) we get

    \(\cot A\cot B + \cot B\cot C + \cot C\cot A = 1\)

    We have already proven above in problem 19.

  31. Let \(A - B = \alpha, B - C = \beta, C - A = \gamma,\) then

    \(\alpha + \beta + \gamma = 0\)

    \(\Rightarrow \tan(\alpha + \beta) = -\tan\gamma\)

    \(\Rightarrow \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = -\tan\gamma\)

    \(\tan\alpha + \tan\beta + \tan\gamma = \tan\alpha\tan\beta\tan\gamma\)

    Substituting back the values, we get

    \(\tan(A - B) + \tan(B - C) + \tan(C - A) = \tan(A - B)\tan(B - C)\tan(C - A)\)

  32. We have already proven in problem 19 that if \(A + B + C = 0,\) then

    \(\cot A\cot B + \cot B\cot C + \cot C\cot A = 1\)

    Let \(A = x + y - z, B = z + x - y, C = y + z - x,\) then

    \(A + B + C = x + y + z = 0\)

    \(\Rightarrow \cot A\cot B + \cot B\cot C + \cot C\cot A = 1\)

    Substituting back the values, we get

    \(\cot(x + y - z)\cot(z + x - y) + \cot(x + y - z)\cot(y + z - x) + \cot(y + z - x)\cot(z + x - y) = 1\)

  33. Given \(A + B + C= n\pi \Rightarrow \tan(A + B) = \tan(n\pi - C) = -\tan C\)

    \(\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan B} = -\tan C\)

    \(\Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C\)

  34. L.H.S \(= (\sin2A + \sin2B) + \sin2C = 2\sin(A + B)\cos(A - B) + \sin2C\)

    \(= 2\sin(\pi - C)\cos(A - B) + \sin2C = 2\sin C\cos(A - B) + 2\sin C\cos C\)

    \(= 2\sin C[\cos(A - B) + \cos\{\pi - (A + B)\}] = 2\sin C[\cos(A - B) - \cos(A + B)]\)

    \(= 4\sin A\sin B\sin C\)

  35. L.H.S. \(= (\cos A + \cos B) + \cos C - 1 = 2\cos\frac{A + B}{2}\cos\frac{A - B}{2} + \cos C - 1\)

    \(= 2\cos\left(\frac{\pi}{2} - \frac{C}{2}\right)\cos\frac{A - B}{2} + \cos C - 1\)

    \(= 2\sin\frac{C}{2}\cos\frac{A - B}{2} + 1 - 2\sin^2\frac{C}{2} - 1\)

    \(= 2\sin\frac{C}{2}\left[\cos\frac{A - B}{2} - \sin\frac{C}{2}\right]\)

    \(= 2\sin\frac{C}{2}\left[\cos\frac{A - B}{2} - \sin\left(\frac{\pi}{2} - \frac{A + B}{2}\right)\right]\)

    \(= 2\sin\frac{C}{2}\left[\cos\frac{A - B}{2} - \cos\frac{A + B}{2}\right]\)

    \(= 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\)

  36. We have proven in 34 and 35 that \(\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C\) and \(\cos A + \cos B + \cos C - 1 = 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\) respectively. Thus,

    \(\frac{\sin 2A + \sin 2B + \sin 2C}{\cos A + \cos B + \cos C - 1} = \frac{4\sin A\sin B\sin C}{4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}\)

    \(= \frac{4.2\sin\frac{A}{2}\cos\frac{A}{2}.2\sin\frac{B}{2}\cos\frac{B}{2}.2\sin\frac{C}{2}\cos\frac{C}{2}}{4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}\)

    \(= 8\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\)

  37. L.H.S. \(= \left(\cos \frac{A}{2} + \cos\frac{B}{2}\right) + \cos\frac{C}{2}\)

    \(= 2\cos\frac{A + B}{4}\cos\frac{A - B}{4} + \sin\frac{\pi - C}{2}\)

    \(= 2\cos\frac{\pi - C}{4}\cos\frac{A - B}{4} + 2\sin\frac{\pi - C}{4}\cos\frac{\pi - C}{4}\)

    \(= 2\cos\frac{\pi - C}{4}\left[\cos\frac{A - B}{4} + \cos\left(\frac{\pi}{2} - \frac{\pi - C}{4}\right)\right]\)

    \(= 2\cos\frac{\pi - C}{4}2\cos\frac{\pi + A + C - B}{8}\cos\frac{\pi + C - A + B}{8}\)

    \(= 4\cos\frac{\pi - A}{4}\cos\frac{\pi - B}{4}\cos\frac{\pi - C}{4}\)

  38. L.H.S. \(= \left(\sin\frac{A}{2} + \sin \frac{B}{2}\right) + \sin\frac{C}{2}\)

    \(= 2\sin\frac{A + B}{4}\cos\frac{A - B}{4} + \cos\frac{\pi - C}{2}\)

    \(= 2\sin\frac{\pi - C}{4}\cos\frac{A - B}{4} + 1 - 2\sin^2\frac{\pi - C}{4}\)

    \(=1 + 2\sin\frac{\pi - C}{4}\left[\cos\frac{A - B}{4} - \sin\frac{\pi - C}{4}\right]\)

    \(= 1 + 2\sin\frac{\pi - C}{4}\left[\cos\frac{A - B}{4} - \cos\frac{\pi + C}{4}\right]\)

    \(= 1 + 2\sin\frac{\pi - C}{4}.2\sin\frac{\pi + A + C - B}{8}\sin\frac{\pi + C - A + B}{8}\)

    \(= 1 + 4\sin \frac{B + C}{4}\sin \frac{C + A}{4}\sin \frac{A + B}{4}\)

  39. L.H.S. \(= \frac{1 - \cos A}{2} + \frac{1 - \cos B}{2} - \frac{1 - \cos C}{2}\)

    \(= \frac{1}{2} - \frac{1}{2}[\cos A + \cos B - \cos C]\)

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

    \(= 2\sin\frac{C}{2}\cos\frac{A - B}{2} - 1 + 2\sin^2\frac{C}{2}\)

    \(= -1 + 2\sin\frac{C}{2}\left[\cos\frac{A - B}{2} + \sin\frac{C}{2}\right]\)

    \(= -1 + 2\sin\frac{C}{2}\left[\cos\frac{A - B}{2} + \cos\frac{A + B}{2}\right]\)

    \(= -1 + 2\sin\frac{C}{2}.2\cos\frac{A}{2}\cos\frac{B}{2}\)

    \(= -1 + 4\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2}\)

    Thus, L.H.S. \(= 1 - 2\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2}\)

  40. L.H.S. \(= 1 + \cos56^\circ + (\cos58^\circ - \cos66^\circ)\)

    \(= 2\cos^228^\circ + 2\sin62^\circ\sin4^\circ\)

    \(=2\cos^228^\circ + 2\cos28^\circ\sin4^\circ\)

    \(= 2\cos28^\circ[\sin4^\circ + \cos28^\circ]\)

    \(= 4\cos28^\circ\cos29^\circ\sin33^\circ\)

  41. Given \(A + B + C = \pi,\) we have to prove that \(\cos 2A + \cos 2B - \cos 2C = 1 - 4\sin A\sin B\cos C\)

    L.H.S. \(=\cos 2A + \cos 2B - \cos 2C = \cos 2A + \cos 2B - \cos[2\pi - 2(A + B)]\)

    \(= 2\cos(A + B)\cos(A - B) - \cos2(A + B) = 2\cos(A + B)\cos(A - B) - 2\cos^2(A + B) + 1\)

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

    \(= 1 - 4\sin A\sin B\cos C[\because\cos(A + B) = \cos(\pi - C) = -\cos C]\)

  42. Given \(A + B + C = \pi,\) we have to prove that \(\sin 2A + \sin 2B - \sin 2C = 4\cos A\cos B\sin C\)

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

    \([\because \sin(A + B) = \sin(\pi - C) = \sin C, \cos C = \cos[\pi - (A + B)] = -\cos(A + B)]\)

    \(=2\sin C[\cos(A - B) + \cos(A + B)]\)

    \(= 4\cos A\cos B\sin C\)

  43. Given \(A + B + C = \pi,\) we have to prove that \(\sin A + \sin B + \sin C = 4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}\)

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

    \(= 2\sin\frac{\pi - C}{2}\cos\frac{A - B}{2} + 2\sin\frac{C}{2}\cos\frac{C}{2}\)

    \(= 2\cos\frac{C}{2}\cos\frac{A - B}{2} + 2\sin\frac{\pi - A - B}{2}\cos\frac{C}{2}\)

    \(= 2\cos\frac{C}{2}[\cos\frac{A - B}{2} + \cos\frac{A + B}{2}]\)

    \(= 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\)

  44. L.H.S. \(= \cos A + \cos B - \cos C = 2\cos\frac{A + B}{2}\cos\frac{A - B}{2} - 1 + 2\sin^2\frac{C}{2}\)

    \(= 2\cos\left(\frac{\pi - C}{2}\right)\cos\frac{A - B}{2} + 2\sin^2\frac{C}{2} - 1\)

    \(= 2\sin\frac{C}{2}\left[\cos\frac{A - B}{2} + \cos\left(\frac{[pi}{2} - \frac{C}{2}\right)\right] - 1\)

    \(= 2\sin\frac{C}{2}\left[\cos\frac{A - B}{2} + \cos\frac{A + B}{2}\right] - 1\)

    \(= 4\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2} - 1\)

  45. \(B + C - A = \pi - A - A = \pi - 2A, C + A - B = \pi - 2B, A + B - C = \pi - 2C\)

    \(\Rightarrow\) L.H.S. \(= \sin 2A + \sin 2B + \sin 2C\)

    We have proven in problem 34 that \(\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C\)

    \(\therefore \sin(B + C - A) + \sin(C + A - B) + \sin(A + B - C) = 4\sin A\sin B\sin C\)

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

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

    \(= \frac{\sin 2A + \sin 2B + \sin 2C}{2\sin A\sin B\sin C}\)

    We have proven in problem 34 that \(\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C\)

    \(\Rightarrow \frac{\sin 2A + \sin 2B + \sin 2C}{2\sin A\sin B\sin C} = 2\)

  47. Given \(A + B + C = \pi,\) we have to prove that \(\frac{\sin 2A + \sin 2B + \sin 2C}{\sin A + \sin B + \sin C} = 8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\)

    We have proven in problem 34 that \(\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C\)

    We have also proven in problem 43 that \(\sin A + \sin B + \sin C = 4\cos \frac{A}{2}\cos \frac{B}{2}\cos\frac{C}{2}\)

    Thus, L.H.S. \(= \frac{4\sin A\sin B\sin C}{4\cos \frac{A}{2}\cos \frac{B}{2}\cos\frac{C}{2}}\)

    \(= 8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\)

  48. Given \(x + y + z = \frac{\pi}{2},\) we have to prove that \(\cos(x - y - z) + \cos(y - z - x) + \cos(z - x - y) - 4\cos x\cos y\cos z = 0\)

    \(x - y - z = x - \frac{\pi}{2} + x = 2x - \frac{\pi}{2}\)

    Similarly \(y - z - x = 2y - \frac{\pi}{2}\) and \(z - x - y = 2z - \frac{\pi}{2}\)

    \(\therefore\) L.H.S. \(= \sin 2x + \sin 2y + \sin 2z - 4\cos x\cos y \cos z\)

    Now, \(\sin 2x + \sin 2y + \sin 2z = 2\sin(x + y)\cos(x - y) + 2\sin z\cos z\)

    \(= 2\cos z\cos(x - y) + 2\sin\left(\frac{\pi}{2} - x - y\right)\cos z\)

    \(= 2\cos z[\cos(x - y) + \cos(x + y)]\)

    \(= 4\cos x\cos y\cos z\)

    \(\therefore \sin 2x + \sin 2y + \sin 2z - 4\cos x\cos y \cos z = 0\)

  49. We have to prove that \(\sin(x - y) + \sin(y - z) + \sin(z - x) + 4\sin\frac{x - y}{2}\sin\frac{y - z}{2}\sin \frac{z - x}{2} = 0\)

    Let \(x - y = \alpha, y - z = \beta\) and \(z - x = \gamma\) then \(\alpha + \beta + \gamma = 0\)

    The given equation becomes \(\sin\alpha + \sin\beta + \sin\gamma + 4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2} = 0\)

    Considering \(\sin\alpha + \sin\beta + \sin\gamma\)

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

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

    \(= -4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\)

    Thus, \(\sin(x - y) + \sin(y - z) + \sin(z - x) + 4\sin\frac{x - y}{2}\sin\frac{y - z}{2}\sin \frac{z - x}{2} = 0\)

  50. \(B + 2C = \pi - A + C, C + 2A = \pi - B + A, A + 2B = \pi - C + B\)

    Thus, L.H.S. \(= -[\sin(C - A) + \sin(A - B) + \sin(B - C)]\)

    Also, note that \(A - B + B - C + C - A = 0\) and we have proven in previous problem that \(\sin\alpha + \sin\beta + \sin\gamma = 4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\) when \(\alpha + \beta + \gamma = 0\)

    Thus, \(\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = 4\sin\frac{B - C}{2}\sin\frac{C - A}{2}\sin\frac{A - B}{2}\)

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

    Following the result of 43 we can say that

    \(\sin\frac{\pi - A}{2} + \sin\frac{\pi - B}{2} + \sin\frac{\pi - C}{2} = 4\cos\frac{\pi - A}{4}\cos\frac{\pi - B}{4}\cos\frac{\pi - C}{4}\)

  52. Let \(x = \tan A, y = \tan B, z = \tan C\)

    Given, \(xy + yz + zx = 1\)

    \(\therefore \tan A\tan B + \tan B\tan C + \tan C\tan A = 1\)

    \(\Rightarrow \tan C(\tan A + \tan B) = 1 - \tan A\tan B\)

    \(\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan B} = \frac{1}{\tan C} = \cot C\)

    \(\Rightarrow \tan(A + B) = \tan\left(\frac{\pi}{2} - C\right)\)

    \(\Rightarrow A + B = \frac{\pi}{2} - C \Rightarrow A + B + C = \frac{\pi}{2}\)

    L.H.S. \(= \frac{x}{1 - x^2} + \frac{y}{1 - y^2} + \frac{z}{1 - z^2}\)

    \(= \frac{\tan A}{1 - \tan^2A} + \frac{\tan B}{1 - \tan^2B} + \frac{\tan C}{1 - \tan^2C}\)

    \(= \frac{1}{2}(\tan 2A + \tan 2B + \tan 2C)\)

    We have already proven that if \(2A + 2B + 2C = \pi\) then \(\tan2A + \tan2B + \tan2C = \tan2A\tan2B\tan2C\)

    \(\therefore \frac{1}{2}(\tan 2A + \tan 2B + \tan 2C) = \frac{1}{2}\tan2A\tan2B\tan2C\)

    \(= \frac{1}{2}\frac{2\tan A}{1 - \tan^2A}.\frac{2\tan B}{1 - \tan^2B}.\frac{2\tan C}{1 - \tan^2C}\)

    \(= \frac{4xyz}{(1 - x^2)(1 - y^2)(1 - z^2)}\)

  53. Let \(x = \tan A, y = \tan B, z = \tan C\)

    Now, \(x + y + z = xyz\)

    \(\Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C\)

    \(\Rightarrow \tan A + \tan B = \tan C(\tan A\tan B - 1)\)

    \(\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan C} = -\tan C = \tan(\pi - C)\)

    \(\Rightarrow A + B = \pi - C \Rightarrow A + B + C = \pi\)

    L.H.S. \(= \frac{3x - x^3}{1 - 3x^2} + \frac{3y - y^3}{1 - 3y^2} + \frac{3z - z^3}{1 - 3z^2}\)

    \(= \frac{3\tan A - \tan^3A}{1 - 3\tan^2A} + \frac{3\tan B - \tan^3B}{1 - 3\tan^2B} + \frac{3\tan C - \tan^3C}{1 - 3\tan^2C}\)

    \(= \tan 3A + \tan 3B + \tan 3C\)

    Now following like prebious problem

    \(\tan3A + \tan3B + \tan3C = \tan3A\tan3B\tan3C\)

    \(= \frac{3\tan A - \tan^3A}{1 - 3\tan^2A}\frac{3\tan B - \tan^3B}{1 - 3\tan^2B}\frac{3\tan C - \tan^3C}{1 - 3\tan^2C}\)

    \(= \frac{3x - x^3}{1 - 3x^2}.\frac{3y - y^3}{1 - 3y^2}.\frac{3z - z^3}{1 - 3z^2}\)

  54. Given \(x + y + z = xyz\), let \(x = \tan A, y = \tan B, z = \tan C\)

    \(\Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C\)

    \(\Rightarrow \tan A + \tan B = \tan C(\tan A\tan B - 1)\)

    \(\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan C} = -\tan C = \tan(\pi - C)\)

    \(\Rightarrow A + B = \pi - C \Rightarrow A + B + C = \pi\)

    L.H.S. \(= \frac{2x}{1 - x^2} + \frac{2y}{1 - y^2} + \frac{2z}{1 - z^2}\)

    \(= \frac{2\tan A}{1 - \tan^2A} + \frac{2\tan B}{1 - \tan^2B} + \frac{2\tan C}{1 - \tan^2C}\)

    \(= \tan 2A + \tan 2B + \tan 2C\)

    Following like problem 52

    \(\tan 2A + \tan 2B + \tan 2C = \tan2A\tan2B\tan2C = \frac{2x}{1 - x^2}.\frac{2y}{1 - y^2}.\frac{2z}{1 - z^2}\)

  55. Given \(x + y + z = xyz\), let \(x = \tan A, y = \tan B, z = \tan C\)

    \(\Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C\)

    \(\Rightarrow \tan A + \tan B = \tan C(\tan A\tan B - 1)\)

    \(\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan C} = -\tan C = \tan(\pi - C)\)

    \(\Rightarrow A + B = \pi - C \Rightarrow A + B + C = \pi\)

    Given, \(x(1 - y^2)(1 - z^2) + y(1 - z^2)(1 - x^2) + z(1 - x^2)(1 - y^2) = 4xyz\)

    Dividing both sides with \((1 - x^2)(1 - y^2)(1 - z^2),\) we get

    \(\frac{x}{1 - x^2} + \frac{y}{1 - y^2} + \frac{z}{1 - z^2} = \frac{4xyz}{(1 - x^2)(1 - y^2)(1 - z^2)}\)

    L.H.S. \(= \frac{x}{1 - x^2} + \frac{y}{1 - y^2} + \frac{z}{1 - z^2} = \frac{1}{2}[\tan 2A + \tan 2B + \tan 2C]\)

    \(= \frac{1}{2}\tan2A\tan2B\tan2C = \frac{4xyz}{(1 - x^2)(1 - y^2)(1 - z^2)}\)

  56. L.H.S. \(= (\cos A + \cos B) + (\cos C + \cos D)\)

    \(= 2\cos\frac{A + B}{2}\cos\frac{A - B}{2} + 2\cos\frac{C + D}{2}\cos\frac{C - D}{2}\)

    \(= 2\cos\frac{A + B}{2}\cos\frac{A - B}{2} + 2\cos\left(\pi - \frac{A + B}{2}\right)\cos\frac{C - D}{2}\)

    \(= 2\cos\frac{A + B}{2}\cos\frac{A - B}{2} - 2\cos\frac{A + B}{2}\cos\frac{C - D}{2}\)

    \(= 2\cos\frac{A + B}{2}\left[\cos\frac{A - B}{2} - \cos\frac{C - D}{2}\right]\)

    \(= 2\cos\frac{A + B}{2}.2\sin\frac{A - B + C - D}{4}\sin\frac{C - D - A + B}{4}\)

    \(= 4\cos\frac{A + B}{2}\sin\frac{A + C - (B + C)}{4}\sin\frac{B + C - (A + D)}{4}\)

    \(= 4\cos\frac{A + B}{2}\sin\frac{A + C -(2\pi - A - C)}{4}\sin\frac{B + C - (2\pi - B - C)}{4}\)

    \(= 4\cos\frac{A + B}{2}\sin\frac{A + C - \pi}{2}\sin\frac{B + C - \pi}{2}\)

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

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

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

    \(= \frac{1}{2}[4 + \{\cos2S + \cos(2S - 2A)\} + \{\cos(2S - 2B) + ]\cos(2S - 2C)\}]\)

    \(= \frac{1}{2}[4 + 2\cos(2S - A)\cos A + 2\cos(2S - B - C)\cos(C - B)]\)

    \(= \frac{1}{2}[4 + 2\cos(B + C)\cos A + 2\cos A\cos(C - B)]\)

    \(= \frac{1}{2}[4 + 2\cos A\{\cos(B + C) + \cos(C - B)\}]\)

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

  58. If \(A + B + C = \pi\) then according to problem 21 \(\tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1\)

    Let \(\tan\frac{A}{2} = x, \tan\frac{B}{2} = y, \tan\frac{C}{2} = z\)

    \(\Rightarrow xy + yz + xz = 1\)

    Now, \((x - y)^2 + (y - z)^2 + (z - x)^2 \geq 0\)

    \(\Rightarrow x^2 + y^2 + z^2 \geq xy + yz + zx\)

    \(\Rightarrow x^2 + y^2 + z^2 \geq 1\)

    \(\tan^2\frac{A}{2} + \tan^2\frac{B}{2} + \tan^2\frac{C}{2}\geq 1\)

  59. We have proven that if \(A + B + C = \pi\) then \(\tan A + \tan B + \tan C = \tan A\tan B\tan C\)

    Thus, L.H.S. \(= \tan A\tan B\tan C(\cot A + \cot B + \cot C)\)

    \(= \tan B\tan C + \tan C\tan A + \tan A\tan B\)

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

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

    \(= \frac{\sin C[\cos A\sin B + \sin A\cos B] + \cos C\sin A\sin B}{\cos A\cos B\cos C}\)

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

    \(= \frac{\sin^2C + \cos C\sin A\sin B}{\cos A\cos B\cos C}\)

    \(= \frac{1 - \cos^2C + \cos C\sin A\sin B}{\cos A\cos B\cos C}\)

    \(= \frac{1 + \cos C[\sin A\sin B - \cos C]}{]\cos A\cos B\cos C}\)

    \(= \frac{1 + \cos C[\sin A\sin B - \cos(\pi - A - B)]}{\cos A\cos B\cos C}\)

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

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

  60. \(\cot B\ + \cot C = \frac{\cos B}{\sin B} + \frac{\cos C}{\sin C}\)

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

    \(= \frac{\sin(\pi - A)}{\sin B\sin C} = \frac{\sin A}{\sin B\sin C}\)

    Similarly, \(\cot C + \cot A = \frac{\sin B}{\sin C\sin A}\) and \(\cot A + \cot B = \frac{\sin C}{\sin A + \sin B}\)

  61. L.H.S. \(= \sin^2A\sin B\cos B + \sin^2A \sin C\cos C + \sin^2B\sin A\cos A + \sin^2B\sin C\cos C + \sin^2C\sin A\cos A + \sin^2C\sin B\cos B\)

    \(= (\sin^2A\sin B\cos B + \sin^2B\sin B\cos B) + (\sin^2A\sin C\cos C + \sin^2C\sin A\cos A) + (\sin^2B\sin C\cos C + \sin^2C\sin B\cos B)\)

    \(= \sin A\sin B\sin(A + B) + \sin A\sin C\sin(A + C) + \sin B\sin C\sin(B + C)\)

    \(= \sin A\sin B\sin C + \sin A\sin C\sin B + \sin B\sin C\sin A\)

    \(= 3\sin A\sin B\sin C\)

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

    \(= 2\sin\frac{A + B}{2}\sin\frac{B - A}{2} + 2\sin\frac{2\pi - (A + B)}{2}\sin\frac{D - C}{2}\)

    \(= 2\sin\frac{A + B}{2}\left[\sin\frac{B - A}{2} + \sin\frac{D - C}{2}\right]\)

    \(= 2\sin\frac{A + B}{2}.2\sin\frac{B + D - (A + C)}{4}\cos\frac{B + C - (A + D)}{4}\)

    \(= 4\sin\frac{A + B}{2}\sin\frac{2\pi - 2(A + C)}{4}\cos\frac{2\pi - 2(A + D)}{4}\)

    \(= 4\sin\frac{A + B}{2}\sin\frac{A + D}{2}\cos \frac{A + C}{2}\)

  63. Since \(A, B, C, D\) are angles of a cyclic quadrilateral \(\therefore A + B + C + D = 2\pi, A + C = \pi, B + D = \pi\)

    We have proven in problem 56 that \(\cos A + \cos B + \cos C + \cos D = 4\cos\frac{A + B}{2}\cos\frac{B + C}{2}\cos\frac{C + A}{2}\)

    \(\cos\frac{A + C}{2} = \cos\frac{\pi}{2} = 0\)

    \(\therefore \cos A + \cos B + \cos C + \cos D = 0\)

  64. We know that \((\cot A - \cot B)^2 + (\cot B - \cot C)^2 + (\cot C - \cot A)^2 \geq 0\)

    Also, \(\because A + B + C = \pi \therefore \cot A\cot B + \cot B\cot C + \cot C\cot A = 1\)

    \(\therefore 2\cot^2A + 2\cot^2B + 2\cot^2C \geq 2(\cot A\cot B + \cot B\cot C + \cot C\cot A)\)

    \(\cot^2A + \cot^2B + \cot^2C \geq 1\)

  65. \(\cos\frac{A}{2}\cos\frac{B - C}{2} = \cos\left(\frac{\pi}{2} - \frac{B + C}{2}\right)\cos\frac{B - C}{2}\)

    \(= \sin\frac{B + C}{2}\cos\frac{B - C}{2} = \frac{1}{2}\left(2\sin\frac{B + C}{2}\cos\frac{B - C}{2}\right)\)

    \(= \frac{1}{2}(\sin B + \sin C)\)

    Similarly \(\cos\frac{B}{2}\cos\frac{C - A}{2} = \frac{1}{2}(\sin A + \sin C)\)

    and \(\cos\frac{C}{2}\cos\frac{A - B}{2} = \frac{1}{2}(\sin A + \sin B)\)

    Adding all these we have desired result.

  66. \(\sin 3A\sin(B - C) = (3\sin A - 4\sin^3A)\sin(B - C)\)

    \(= 3\sin A\sin(B - C) - 4\sin^2A.\sin A\sin(B - C)\)

    \(= 3\sin(B + C)\sin(B - C) - 4\sin^2A\sin(B + C)\sin(B - C)[\because B + C = \pi - A \Rightarrow \sin(B + C) = \sin A]\)

    \(= \frac{3}{2}(\cos2C - \cos2B) - 2\sin^2A(\cos2C - \cos2B)\)

    Now, \(2\sin^2A(\cos2C - \cos2B) = (1 - \cos 2A)(\cos2C - \cos2B)\)

    \(= \cos 2C - \cos 2B - \cos2C \cos 2A + \cos 2A\cos 2B\)

    Thus, \(\sin 3A\sin(B - C) + \sin 3B\sin(C - A) + \sin3C\sin(A - B) = 0\)