# 15. Trigonometrical Identities Solutions¶

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