# 13. Multiple and Submultiple Angles Solutions¶

1. Let us solve these one by one.

1. Given, $$\cos A = \frac{3}{5}$$

$$\Rightarrow \sin A = \sqrt{1 - \cos^2A} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$

$$\sin 2A = 2\sin A\cos A = 2.\frac{4}{5}.\frac{3}{5} = \frac{24}{25}$$

2. Given, $$\sin A = \frac{12}{13}$$

$$\Rightarrow \cos A = \sqrt{1 - \sin^2A} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$$

$$\sin 2A = 2\sin A\cos A = 2.\frac{12}{13}.\frac{5}{13} = \frac{120}{169}$$

3. Given, $$\tan A = \frac{16}{63} = \frac{\text{perpendicular}}{\text{base}}$$

$$\text{hypotenuse} = \sqrt{p^2 + b^2} = \sqrt{16^2 + 63^2} = 65$$

$$\sin A = \frac{16}{65}, \cos A = \frac{63}{65}$$

$$\sin 2A = 2\sin A\cos A = 2.\frac{16}{65}.\frac{63}{65} = \frac{2016}{4225}$$

2. Let us solve these one by one.

1. Given, $$\cos A = \frac{15}{17}$$

$$\Rightarrow \sin A = \sqrt{1 - \cos^2A}$$ $$= \sqrt{1 - \frac{225}{289}} = \sqrt{\frac{64}{289}} = \frac{8}{17}$$

$$\cos 2A = \cos^2A - \sin^2A = \frac{225 - 64}{289} = \frac{161}{289}$$

2. Given, $$\sin A = \frac{4}{5}$$

$$\Rightarrow \cos A = \sqrt{1 - \sin^2A}$$ $$= \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$$

$$\cos2A = \cos^2A - \sin^2A = \frac{9 - 16}{25} = -\frac{7}{25}$$

3. Give, $$\tan A = \frac{5}{12} = \frac{\text{perpendicular}}{\text{base}}$$

$$\text{hypotenuse} = \sqrt{p^2 + b^2} = \sqrt{25 + 144} = 13$$

$$\sin A = \frac{5}{13}, \cos A = \frac{12}{13}$$

$$\cos^2A = \cos^2A - \sin^2A = \frac{119}{169}$$

3. Given, $$\tan A = \frac{b}{a},$$ thus $$\text{hypotenuse} = \sqrt{b^2 + a^2}$$

$$a\cos 2A+ b\sin 2A = a(\cos^2A - \sin^2A) + 2b\sin A\cos A$$

$$= a\left(\frac{a^2}{a^2 + b^2} - \frac{b^2}{a^2 + b^2}\right) + 2b.\frac{ab}{a^2 + b^2}$$

$$= a\left(\frac{a^2 - b^2 + 2b^2}{a^2 + b^2}\right) = a$$

4. We have to prove that $$\frac{\sin 2A}{1 + \cos 2A} = \tan A$$

L.H.S. $$= \frac{\sin 2A}{1 + \cos 2A} = \frac{2\sin A\cos A}{1 + \cos^2A - \sin^2A}$$

$$= \frac{2\sin A\cos A}{2\cos^2A}[\because 1 - \sin^2A = \cos^2A]$$

$$= \tan A =$$ R.H.S.

5. We have to prove that $$\frac{\sin 2A}{1 - \cos 2A} = \cot A$$

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

$$= \frac{2\sin A\cos A}{2\sin^2A} = \cot A =$$ R.H.S.

6. We have to prove that $$\frac{1 - \cos 2A}{1 + \cos 2A} = \tan^2A$$

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

$$= \frac{2\sin^2A}{2\cos^2A} = \tan^2A =$$ R.H.S.

7. We have to prove that $$\tan A + \cot A = 2\cosec 2A$$

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

$$= \frac{2}{2\sin A\cos A} = \frac{2}{\sin 2A} = 2\cosec 2A =$$ R.H.S.

8. We have to prove that $$\tan A - \cot A = -2\cot2A$$

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

$$= \frac{-\cos2A}{\frac{\sin2A}{2}} = -2\cot2A =$$ R.H.S.

9. We have to prove that $$\cosec 2A + \cot 2A = \cot A$$

L.H.S. $$= \frac{1}{\sin 2A} + \frac{\cos 2A}{\sin 2A} = \frac{1 + \cos 2A}{\sin 2A} = \frac{2\cos^2A}{2\sin A\cos A}$$

$$= \cot A =$$ R.H.S.

10. We have to prove that $$\frac{1 - \cos A + \cos B - \cos(A + B)}{1 + \cos A - \cos B - \cos(A + B)} = \tan\frac{A}{2}\cot\frac{B}{2}$$

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

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

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

$$= \frac{\tan\frac{A}{2}\left(2\sin\left(\frac{A + B}{2}\right)\cos\frac{B}{2}\right)}{2\sin\left(\frac{A + B}{2}\right)\sin\frac{B}{2}}$$

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

11. We have to prove that $$\frac{\cos A}{1 \mp \sin A} = \tan\left(45^\circ \pm \frac{A}{2}\right)$$

First considering $$-$$ sign on L.H.S.,

L.H.S. $$= \frac{\cos A}{1 - \sin A} = \frac{\cos^2\frac{A}{2} - \sin^2\frac{A}{2}}{\left(\cos\frac{A}{2} - \sin\frac{A}{2}\right)^2}$$

Dividing numerator and denomiator by $$\cos^2\frac{A}{2}$$

$$= \frac{1 - \tan^2\frac{A}{2}}{\left(1 - \tan\frac{A}{2}\right)^2}$$

$$= \frac{1 + \tan\frac{A}{2}}{1 - \tan\frac{A}{2}}$$

$$= \frac{\tan45^\circ + \tan\frac{A}{2}}{1 - \tan45^\circ\tan\frac{A}{2}} = \tan\left(45^\circ + \frac{A}{2}\right)$$

Similarly by considering the $$+$$ sign we can prove the other sign.

12. We have to prove that $$\frac{\sec 8A - 1}{\sec 4A - 1} = \frac{\tan 8A}{\tan 2A}$$

L.H.S. $$= \frac{\sec 8A - 1}{\sec 4A - 1} = \frac{1 - \cos 8A}{1 - \cos 4A}.\frac{\cos4A}{\cos8A}$$

$$= \frac{2\sin^24A}{2\sin^22A}.\frac{\cos 4A}{\cos8A} = \frac{(2\sin4A\cos4A).\sin4A}{2\sin^22A.\cos8A}$$

$$= \frac{\sin8A}{\cos8A}.\frac{\sin4A}{2\sin^22A} = \frac{\tan8A. 2\sin2A\cos2A}{2\sin^22A} = \frac{\tan8A}{\tan2A} =$$ R.H.S.

13. We have to prove that $$\frac{1 + \tan^2(45^\circ - A)}{1 - \tan^2(45^\circ - A)} = \cosec 2A$$

L.H.S. $$= \frac{1 + \tan^2(45^\circ - A)}{1 - \tan^2(45^\circ - A)}$$

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

$$= \frac{1}{\cos(90^\circ - 2A)} = \frac{1}{\sin2A} = \cosec2A =$$ R.H.S.

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

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

$$= \frac{\tan \frac{A + B}{2}}{\tan \frac{A - B}{2}} =$$ R.H.S.

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

L.H.S. $$= \frac{2(\cos^2B - \cos^2A)}{\sin2A - \sin2B} = \frac{\cos2B - \cos2A}{\sin2A - \sin2B}$$

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

16. We have to prove that $$\tan\left(\frac{\pi}{4} + A\right) - \tan\left(\frac{\pi}{4} - A\right) = 2\tan 2A$$

L.H.S. $$= \frac{1 + \tan A}{1 - \tan A} - \frac{1 - \tan A}{1 + \tan A}$$

$$= \frac{(1 + \tan A)^2 - (1 - \tan A)^2}{1 - \tan^2A} = \frac{4\tan A}{1 - \tan^2A}$$

$$= \frac{4\sin A}{\cos A}. \frac{\cos^2A}{\cos^2A - \sin^2A} = \frac{2\sin2A}{\cos2A} = 2\tan2A =$$ R.H.S.

17. We have to prove that $$\frac{\cos A + \sin A}{\cos A - \sin A} - \frac{\cos A - \sin A}{\cos A + \sin A} = 2\tan 2A$$

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

$$= \frac{4\cos A\sin A}{\cos 2A} = \frac{2\sin 2A}{\cos 2A} = 2\tan2A =$$ R.H.S.

18. We have to prove that $$\cot (A + 15^\circ) - \tan(A - 15^\circ) = \frac{4\cos 2A}{1 + 2\sin 2A}$$

L.H.S. $$= \frac{1 - \tan(A + 15^\circ)\tan(A - 15^\circ)}{\tan(A + 15^\circ)}$$

$$= \frac{\cos(A + 15^\circ)\cos(A - 15^\circ) - \sin(A + 15^\circ)\sin(A - 15^\circ)}{\cos(A + 15^\circ)\cos(A - 15^\circ)}.\frac{\cos(A + 15^\circ)}{\sin(A + 15^\circ)}$$

$$= \frac{\cos 2A}{\sin(A + 15^\circ)\cos(A - 15^\circ)} = \frac{2\cos 2A}{2\sin(A + 15^\circ)\cos(A - 15^\circ)}$$

$$= \frac{2\cos 2A}{\sin2A + \sin30^\circ} = \frac{4\cos 2A}{1 + \sin 2A} =$$ R.H.S.

19. We have to prove that $$\frac{\sin A + \sin2A}{1 + \cos A + \cos 2A} = \tan A$$

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

$$= \tan A =$$ R.H.S.

20. We have to prove that $$\frac{1 + \sin A - \cos A }{1 + \sin A + cos A} = \tan \frac{A}{2}$$

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

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

$$= \tan\frac{A}{2} =$$ R.H.S.

21. We have to prove that $$\frac{\sin(n + 1)A - \sin(n - 1)A}{\cos(n + 1)A + 2\cos nA + \cos(n - 1)A} = \tan \frac{A}{2}$$

L.H.S. $$= \frac{2\cos nA \sin A}{2\cos nA \cos A + 2\cos nA} = \frac{\sin A}{1 + \cos A}$$

$$= \frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{2\cos^2\frac{A}{2}} = \tan \frac{A}{2} =$$ R.H.S.

22. We have to prove that $$\frac{\sin(n + 1)A + 2\sin nA + \sin(n - 1)A}{\cos(n - 1) - \cos(n + 1)A} = \cot \frac{A}{2}$$

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

$$= \frac{\cos A + 1}{\sin A} = \frac{2\cos^2\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}}$$

$$= \cot\frac{A}{2} =$$ R.H.S.

23. We have to prove that $$\sin(2n + 1)A\sin A = \sin^2(n + 1)A - \sin^2nA$$

R.H.S. $$= (\sin(n + 1)A + \sin nA)(\sin(n + 1)A - \sin nA)$$

$$= (2\sin\frac{2n + 1}{2}A\cos \frac{A}{2})(2\cos\frac{2n + 1}{2}A\sin \frac{A}{2})$$

$$= 2\sin\frac{2n + 1}{2}A\cos\frac{2n + 1}{2}A.2\cos \frac{A}{2}\sin\frac{A}{2}$$

$$= \sin(2n + 1)A\sin A =$$ L.H.S.

24. We have to prove that $$\frac{\sin(A + 3B) + \sin(3A + B)}{\sin 2A + \sin 2B} = 2\cos(A + B)$$

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

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

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

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

L.H.S. $$= 2\cos 2A\sin A + 2\sin A\cos A = 2\sin A(\cos 2A + \cos A)$$

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

26. We have to prove that $$\tan 2A = (\sec 2A + 1)\sqrt{\sec^2A - 1}$$

R.H.S. $$= \frac{1 + \cos 2A}{\cos 2A}\sqrt{\frac{1 - \cos^2A}{\cos^2A}}$$

$$= \frac{2\cos^2A}{2\cos^2A - 1}.\sqrt{\frac{\sin^2A}{\cos^2A}}$$

$$= \frac{2}{2 - \sec^2A}.tan A = \frac{2\tan A}{1 - \tan^2A} = \frac{\tan A + \tan A}{1 - \tan A.\tan A}$$

$$=\tan 2A =$$ R.H.S.

27. We have to prove that $$\cos^32A + 3\cos 2A = 4(\cos^6A - \sin^6A)$$

L.H.S. $$= (\cos^2A - \sin^2A)^3 + 3(\cos^2A - \sin^2A)$$

$$= \cos^6A -3\cos^4A\sin^2A + 3\cos^2A\sin^4A - \sin^6A + 3(\cos^2A - \sin^2A)$$

$$= \cos^6A -3\cos^4A(1 - \cos^2A) + 3(1 - \sin^2A)\sin^4A - \sin^6A + 3(\cos^2A - \sin^2A)$$

$$= 4(\cos^6A - \sin^6A) =$$ R.H.S.

28. We have to prove that $$1 + \cos^22A = 2(\cos^4A + \sin^4A)$$

L.H.S. $$= 1 + (\cos^2A - \sin^2A)^2 = 1 - 2\sin^2A\cos^2A + \cos^4A + \sin^4A$$

$$= 1 - 2\sin^2A(1 - \sin^2A) + \cos^4A + \sin^4A$$

$$= 1 - 2\sin^2A + 2\sin^4A + \cos^4A + \sin^4A$$

$$= (1 - \sin^2A)^2 + \cos^4A + 2\sin^4A = 2(\cos^4A + \sin^4A) =$$ R.H.S.

29. We have to prove that $$\sec^2A(1 + \sec2A) = 2\sec2A$$

L.H.S. $$= \frac{1}{\cos^2A}.\frac{\cos2A + 1}{\cos 2A}$$

$$= \frac{1}{\cos^2A}.\frac{2\cos^2A}{\cos 2A} = 2\sec2A =$$ R.H.S.

30. We have to prove that $$\cosec A - 2\cot 2A\cos A = 2\sin A$$

L.H.S. $$= \frac{1}{\sin A} - \frac{2\cos 2A\cos A}{\sin 2A}$$

$$= \frac{1}{\sin A} - \frac{2\cos 2A\cos A}{2\sin A\cos A}$$

$$\frac{1}{\sin A} - \frac{\cos 2A}{\sin A} = \frac{1 - \cos 2A}{\sin A}$$

$$= \frac{2\sin^2A}{\sin A} = 2\sin A =$$ R.H.S.

31. We have to prove that $$\cot A = \frac{1}{2}\left(\cot\frac{A}{2} - \tan\frac{A}{2}\right)$$

R.H.S. $$= \frac{1}{2}\left(\frac{1 - \tan^2\frac{A}{2}}{\tan\frac{A}{2}}\right)$$

$$= \frac{1}{2}\left(\frac{\cos^2\frac{A}{2} - \sin^2\frac{A}{2}}{\cos^2\frac{A}{2}}\right).\frac{\cos\frac{A}{2}}{\sin \frac{A}{2}}$$

$$= \frac{1}{2}\frac{\cos A}{\cos\frac{A}{2}}.\frac{1}{\sin\frac{A}{2}} = \cot A =$$ L.H.S.

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

L.H.S. $$=\sin A.\frac{\cos 2A - \cos 120^\circ}{2} = \frac{\sin A\left(1 - 2\sin^2A + \frac{1}{2}\right)}{2}$$

$$= \frac{3\sin A - 4\sin^3A}{4} = \frac{1}{4}\sin 3A =$$ R.H.S.

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

L.H.S. $$= \frac{\cos A}{2}\left(\cos 2A + \cos120^\circ\right) = \frac{\cos A}{2}\left(2\cos^2A - 1 - \frac{1}{2}\right)$$

$$= \frac{4\cos^3A - 3\cos A}{4} = \frac{1}{4}\cos 3A =$$ R.H.S.

34. We have to prove that $$\cot A + \cot(60^\circ + A) - \cot(60^\circ - A) = 3\cot 3A$$

L.H.S. $$= \frac{1}{\tan A} + \frac{1}{\tan(60^\circ + A)} - \frac{1}{\tan(60^\circ - A)}$$

$$= \frac{1}{\tan A} + \frac{1 - \sqrt{3}\tan A}{\sqrt{3} + \tan A} - \frac{1 + \sqrt{3}\tan A}{\sqrt{3} - \tan A}$$

$$= \frac{1}{\tan A} - \frac{8\tan A}{3 - \tan^2A} = \frac{3(1 - 3\tan^2A)}{3\tan A - \tan^3A} = \frac{3}{\tan 3A}$$

$$= 3\cot 3A =$$ R.H.S.

35. We have to prove that $$\cos 4A = 1 - 8\cos^2A + 8\cos^4A$$

L.H.S. $$= \cos 4A = 2\cos^22A - 1 = 2(2\cos^2A - 1)^2 - 1$$

$$=2(4\cos^4A - 4\cos^2A + 1) - 1$$

$$= 1 - 8\cos^2A + 8\cos^4A =$$ R.H.S.

36. We have to prove that $$\sin 4A = 4\sin A\cos^3A - 4\cos A\sin^3A$$

L.H.S. $$= 2\sin 2A\cos 2A = 4\sin A\cos A(\cos^2A - \sin^2A)$$

$$= 4\sin A\cos^3A - 4\cos A\sin^3A =$$ R.H.S.

37. We have to prove that $$\cos 6A = 32\cos^6A - 48\cos^4A + 18\cos^2A - 1$$

L.H.S. $$= \cos 6A = (\cos^23A - \sin^23A) = (4\cos^3A - 3\cos A)^2 - (3\sin A - 4\sin^3A)^2$$

$$= 16\cos^6A + 9\cos^2A -24\cos^4A - 9\sin^2A - 16\sin^6A + 24\sin^4A$$

$$= 16\cos^6A + 9\cos^2A -24\cos^4A - 9(1 - \cos^2A) - 16(1 - \cos^2A)^3 + 24(1 - \cos^2A)^2$$

$$= 32\cos^6A - 48\cos^4A + 18\cos^2A - 1 =$$ R.H.S.

38. We have to prove that $$\tan 3A\tan 2A\tan A = \tan 3A - \tan 2A - \tan A$$

Rewriting this as following:

$$\tan A + \tan 2A = \tan 3A(1 - \tan A\tan 2A)\Rightarrow \frac{\tan A + \tan 2A}{1 - \tan A\tan 2A} = \tan 3A$$

$$\Rightarrow \tan (A + 2A) = \tan 3A$$

Hence, proved.

39. We have to prove that $$\frac{2\cos2^nA + 1}{2\cos A + 1} = (2\cos A - 1)(2\cos 2A - 1)(2\cos2^2A - 1)\ldots(2\cos2^{n - 1} - 1)$$

L.H.S. $$= \frac{2\cos2^nA + 1}{2\cos A + 1}$$

Multiplying and dividing by $$2\cos A - 1$$

$$= (2\cos A - 1)\frac{2\cos2^nA + 1}{4\cos^2A - 1} = (2\cos A - 1)\frac{2\cos2^nA + 1}{2\cos2A + 1}$$

Multiplying and dividing by $$2\cos2A - 1$$

$$= (2\cos A - 1)(2\cos2A - 1)\frac{2\cos2^nA + 1}{4\cos^22A - 1}$$

$$= (2\cos A - 1)(2\cos2A - 1)\frac{2\cos2^nA + 1}{2\cos2^2A + 1}$$

Proceeding similarly we obtain the R.H.S.

40. Given $$\tan A= \frac{1}{7}, \sin B = \frac{1}{\sqrt{10}}$$

$$\therefore \cos B = \frac{3}{\sqrt{10}}, \tan B = \frac{1}{3}$$

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

$$= \frac{\tan A + \frac{2\tan B}{1 - \tan^2B}}{1 - \tan A.\frac{2\tan B}{1 - \tan^2B}}$$

$$= \frac{\frac{1}{7} + \frac{2\frac{1}{3}}{1 - \frac{1}{9}}}{1 - \frac{1}{7}.\frac{2\frac{1}{3}}{1 - \frac{1}{9}}}$$

$$= 1 \therefore A + 2B = \frac{\pi}{4}$$

41. We have to prove that $$\tan\left(\frac{\pi}{4} + A\right) + \tan\left(\frac{\pi}{4} - A\right) = 2\sec2A$$

L.H.S. $$= \frac{1 + \tan A}{1 - \tan A} + \frac{1 - \tan A}{1 + \tan A}$$

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

$$= \frac{2(\sin^2A + \cos^2A)}{\cos^2A - \sin^2A} = \frac{2}{\cos 2A} = 2\sec 2A =$$ R.H.S.

42. We have to prove that $$\sqrt{3}\cosec 20^\circ - \sec 20^\circ = 4$$

L.H.S. $$= \frac{\sqrt{3}}{\sin20^\circ} - \frac{1}{\cos20^\circ}$$

$$= \frac{4(\frac{\sqrt{3}}{2})\cos20^\circ - \frac{1}{2}\sin20^\circ}{2\sin20^\circ\cos^20\circ}$$

$$= \frac{4(\sin(50^\circ - 20^\circ))}{\sin40^\circ} = 4 =$$ R.H.S.

43. We have to prove that $$\tan A + 2\tan 2A + 4\tan 4A + 8\cot 8A = \cot A$$

$$\tan A - \cot A = \frac{\sin^2A - \cos^2A}{\sin A\cos A} = -\frac{2\cos 2A}{\sin 2A} = -2\cot 2A$$

Similarly, $$2\tan 2A - 2\cot 2A = -4\cot 4A$$

and $$4\tan 4A - 4\cot 4A = -8\cot 8A$$

Thus, $$\tan A + 2\tan 2A + 4\tan 4A + 8\cot 8A = \cot A$$

44. We have to prove that $$\cos^2A + \cos^2\left(\frac{2\pi}{3} - A\right) + \cos^2\left(\frac{2\pi}{3} + A\right) = \frac{3}{2}$$

$$\Rightarrow 2\cos^2A + 2\cos^2\left(\frac{2\pi}{3} - A\right) + 2\cos^2\left(\frac{2\pi}{3} + A\right) = 3$$

L.H.S. $$= \cos 2A + 1 + \cos\left(\frac{4\pi}{3} - 2A\right) + 1 + \cos\left(\frac{4\pi}{3} + 2A\right) + 1$$

$$= 3 + \cos2A + 2\cos\left(\frac{4\pi}{3}\right)\cos2A = 3 =$$ R.H.S.

45. $$2\sin^2A + 4\cos (A + B)\sin A\sin B + \cos2(A + B)$$

$$= 2\sin^2A + 2\cos(A + B)2\sin A\sin B + \cos2(A + B)$$

$$= 2\sin^2A + 2\cos(A + B)[\cos(A - B) - \cos(A + B)] + \cos2(A + B)$$

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

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

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

46. Given, $$\cos A = \frac{1}{2}\left(a + \frac{1}{a}\right)$$

$$\cos 2A = 2\cos^2A - 1 = 2.\frac{1}{4}\left(a + \frac{1}{a}\right)^2 - 1$$

$$= \frac{1}{2}\left(a^2 + \frac{1}{a^2}\right)$$

47. We have to prove that $$\cos^2A + \sin^2A\cos 2B = \cos^2B + \sin^2B\cos 2A$$

$$\Rightarrow \cos^2A - \cos^2B = \sin^2B\cos2A - \sin^2A\cos2B$$

R.H.S. $$= \sin^2B\cos2A - \sin^2A\cos2B$$

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

$$= \cos^2A\sin^2B - \sin^2A\cos^2B = \cos^2A(1 - \cos^2B) - (1 - \cos^2A)\cos^2B$$

$$= \cos^2A - \cos^2B =$$ R.H.S.

48. We have to prove that $$1 + \tan A\tan 2A = \sec 2A$$

L.H.S. $$= 1 + \tan A\tan 2A = 1 + \tan A.\frac{2\tan A}{1 - \tan^2A}$$

$$= \frac{1 + \tan^2A}{1 - \tan^2A} = \frac{\cos^2A + \sin^2A}{\cos^2A - \sin^2A}$$

$$= \frac{1}{\cos 2A} = \sec 2A =$$ R.H.S.

49. We have to prove that $$\frac{1 + \sin 2A}{1 - \sin 2A} = \left(\frac{1 + \tan A}{1 - \tan A}\right)^2$$

L.H.S. $$= \frac{1 + \sin 2A}{1 - \sin 2A} = \frac{\sin^2A + \cos^2A + 2\sin A\cos A}{\sin^2A + \cos^2A - 2\sin A\cos A}$$

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

Dividing numerator and denominator by $$\cos^2A,$$ we get

$$= \left(\frac{1 + \tan A}{1 - \tan A}\right)^2 =$$ R.H.S.

50. We have to prove that $$\frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ} = 4$$

L.H.S. $$= \frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ}$$

$$= \frac{\cos10^\circ - \sqrt{3}\sin10^\circ}{\sin10^\circ\cos10^\circ}$$

$$= \frac{2.2\left(\frac{1}{2}\cos10^\circ - \frac{\sqrt{3}}{2}\sin10^\circ\right)}{2\sin10^\circ\cos10^\circ}$$

$$= 4.\frac{\sin30^\circ\cos10^\circ - \cos30^\circ\sin10^\circ}{\sin20^\circ} = 4.\frac{\sin(30^\circ - 10^\circ)}{\sin20^\circ}$$

$$= 4 =$$ R.H.S.

51. We have to prove that $$\cot^2A - \tan^2A = 4\cot2A\cosec 2A$$

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

$$= \frac{\cos^4A - \sin^4A}{\sin^2A\cos^2A} = \frac{4(\cos^2A + \sin^2A)(\cos^2A - \sin^2AA)}{(2\sin A\cos A)^2}$$

$$= \frac{4\cos 2A}{\sin^22A} = 4\cot 2A\cosec 2A =$$ R.H.S.

52. We have to prove that $$\frac{1 +\sin 2A}{\cos2A} = \frac{\cos A + \sin A}{\cos A - \sin A} = \tan\left(\frac{\pi}{4} + A\right)$$

L.H.S. $$= \frac{1 +\sin 2A}{\cos2A} = \frac{\sin^2A + \cos^2A + 2\sin A\cos A}{\cos^2A - \sin^2A}$$

$$= \frac{(\cos A + \sin A)^2}{\cos^2A - \sin^2A} = \frac{\cos A + \sin A}{\cos A - \sin A} =$$ middle term

Dividing both numerator and denominator by $$\cos A,$$ we get

$$= \frac{1 + \tan A}{1 - \tan A} = \frac{\tan\frac{\pi}{4} + \tan A}{1 - \tan\frac{\pi}{4}.\tan A}$$

$$= \tan\left(\frac{\pi}{4} + A\right) =$$ R.H.S.

53. We have to prove that $$\cos^6A - \sin^6A = \cos2A\left(1 - \frac{1}{4}\sin^22A\right)$$

R.H.S. $$= \cos2A\left(1 - \frac{1}{4}\sin^22A\right) = (\cos^2A - \sin^2A)(1 - \sin^2A\cos^2A)$$

$$= (\cos^2A - \sin^2A)[(\cos^2A + \sin^2A)^2 - \sin^2A\cos^2A] = \cos^6A - \sin^6A =$$ L.H.S.

54. This problem is similar to 44 and can be solved similarly.

55. We have to prove that $$(1 + \sec2A)(1+ \sec2^2A)(1 + sec2^3A) \ldots (1 + \sec2^nA) = \frac{\tan2^nA}{\tan A}$$

L.H.S. $$= (1 + \sec2A)(1 + \sec2^2A)(1 + \sec2^3A) \ldots (1 + \sec2^nA)$$

$$= \frac{\tan A}{\tan A}(1 + \sec2A)(1 + \sec2^2A)(1 + \sec2^3A) \ldots (1 + \sec2^nA)$$

Now $$\tan A(1 + \sec 2A) = \tan A\frac{1 + \cos 2A}{\cos 2A}$$

$$= \tan A\frac{1 + \frac{1 - \tan^2A}{1 + \tan^2A}}{\frac{1 - \tan^2A}{1 + \tan^2A}}$$

$$= \tan A\frac{2}{1 - \tan^2A} = \frac{2\tan A}{1 - \tan^2A} = \tan 2A$$

Similarly, $$\tan 2A(1 + \sec2^2A) = \tan2^2A$$

Proceeding similalry we obtain R.H.S.

56. We have to prove that $$\frac{\sin2^nA}{\sin A} = 2^n\cos A\cos 2A\cos 2^2A\ldots\cos2^{n - 1}A$$

Dividing and multiplying with $$2\cos A$$

L.H.S. $$= \frac{\sin2^nA}{\sin A} = 2\cos A.\frac{\sin2^nA}{2\sin A\cos A} = 2\cos A.\frac{\sin2^nA}{\sin 2A}$$

Again, dividing and multiplying with $$2\cos 2A$$

$$= 2^2\cos A\cos 2A.\frac{\sin2^n A}{2\sin 2A\cos 2A} = 2^2\cos A\cos 2A.\frac{\sin2^n A}{\sin 2^2A}$$

Proceeding similarly, we find the R.H.S.

57. We have to prove that $$3(\sin A - \cos A)^4 + 6(\sin A + \cos A)^2 + 4(\sin^6A + \cos^6A) = 13$$

$$3(\sin A - \cos A)^4 = 3[(\sin A - \cos A)^2]^2 = 3(1 - \sin 2A)^2$$

$$6(\sin A + \cos A)^2 = 6(1 + \sin2A)$$

$$4(\sin^6A + \cos^6A) = 4[(\cos^2A + \sin^2A)^3 - 3\cos^2A\sin^2A(\cos^2A + \sin^2A)] = 4(1 - \frac{3}{4}\sin^22A)$$

Adding all these yields $$13.$$

58. We have to prove that $$2(\sin^6A + \cos^6A) - 3(\sin^4A + \cos^4A) + 1 = 0$$

L.H.S. $$= 2[(\sin^2A + \cos^2A)^3 - 3\sin^2A\cos^2A(\sin^2A + \cos^2A)] -3[(\sin^2A + \cos^2A)^2 - 2\sin^2A\cos^2A] + 1$$

$$= 2(1 - 3\sin^2A\cos^2A) - 3[1 - 2\sin^2A\cos^2A] + 1 = 0 =$$ R.H.S.

59. Given $$\cos^2A + \cos^2(A + B) -2\cos A\cos B\cos(A + B)$$

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

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

$$= \cos^2A + \sin^2A\sin^2B - \cos^2A\cos^2B$$

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

$$= 1 - \cos^2B$$ which is independent of $$A$$

60. We have to prove that $$\cos^3A\cos 3A + \sin^3A\sin 3A = \cos^32A$$

We know that $$\cos^3A = \frac{1}{4}(3\cos A + \cos 3A)$$ and

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

L.H.S. $$= \frac{1}{4}(3\cos A + \cos 3A)\cos 3A + \frac{1}{4}(3\sin A - \sin 3A)\sin 3A$$

$$= \frac{3}{4}(\cos3A\cos A + \sin 3A\sin A) + \frac{1}{4}(\cos^23A - \sin^23A)$$

$$= \frac{3}{4}\cos 2A + \frac{1}{4}\cos6A$$

$$= \frac{3}{4}\cos 2A + \frac{1}{4}(4\cos^32A - 3\cos 2A)$$

$$= \cos^32A =$$ R.H.S.

61. We have to prove that $$\tan A\tan(60^\circ - A)\tan(60^\circ + A) = \tan 3A$$

L.H.S. $$= \frac{\sin A.\sin(60^\circ - A).\sin(60^\circ + A)}{\cos A.\cos(60^\circ - A).\cos(60^\circ + A)}$$

$$= \frac{\sin A(\sin^260^\circ - \sin^2A)}{\cos A(\cos^260^\circ - \sin^2A)}[\because \sin(A + B)\sin (A - B) = \sin^2A - \sin^2B\text{~and~}\cos(A + B)\cos(A - B) = \cos^2A - \sin^2B]$$

$$= \frac{\sin A(3 - 4\sin^2A)}{\cos A(1 - 4\sin^2A)} = \frac{3\sin A - 4\sin^3A}{4\cos^3A - 3\cos A}$$

$$= \frac{\sin 3A}{\cos 3A} = \tan 3A =$$ R.H.S.

62. We have to prove that $$\sin^2A + \sin^3\left(\frac{2\pi}{3} + A\right) + \sin^3\left(\frac{4\pi}{3} + A\right) = -\frac{3}{4}\sin 3A$$

$$\because \sin^3A = \frac{1}{4}[3\sin A - \sin 3A]$$

L.H.S. $$= \frac{1}{4}[3\sin A - \sin 3A] + \frac{1}{4}\left[3\sin\left(\frac{2\pi}{3} + A\right) - \sin(2\pi + 3A)\right] + \frac{1}{4}\left[3\sin\left(\frac{4\pi}{3} + A\right) - \sin(4\pi + 3A)\right]$$

$$= \frac{1}{4}[3\sin A - \sin 3A] + \frac{1}{4}\left[3\sin\left(\frac{2\pi}{3} + A\right) - \sin3A\right] + \frac{1}{4}\left[3\sin\left(\frac{4\pi}{3} + A\right) - \sin3A\right]$$

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

$$= \frac{3}{4}[\sin A - \sin 3A + 2.(-\sin A).\frac{1}{2}]$$

$$= -\frac{3}{4}\sin 3A =$$ R.H.S.

63. We have to prove that $$4(\cos^310^\circ + \sin^320^\circ) = 3(\cos 10\circ + \sin 20^\circ)$$

$$\Rightarrow 4\cos^310^\circ - 3\cos10^\circ = 3\sin20^\circ - 4\sin^320^\circ$$

$$\Rightarrow \cos 3.10^\circ = \sin 3.20^\circ$$

$$\Rightarrow \cos 30^\circ = \sin 60^\circ \Rightarrow \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$

Hence, proved.

64. We have to prove that $$\sin A\cos^3A - \cos A\sin^3A = \frac{1}{4}\sin 4A$$

L.H.S. $$= \frac{1}{2}2\sin A\cos A(\cos^2A - \sin^2A) = \frac{1}{2}\sin2A\cos 2A$$

$$= \frac{1}{4}.2.\sin2A\cos 2A = \frac{1}{4}\sin 4A =$$ R.H.S.

65. We have to prove that $$\cos^3A\sin3A + \sin^3A\cos 3A = \frac{3}{4}\sin 4A$$

L.H.S. $$= \cos^3A(3\sin A - 4\sin^3A) + \sin^3A(4\cos^3A - 4\cos A)$$

$$= 3(\sin A\cos^3A - \cos A\sin^3A)$$

Following previous problem we obtain R.H.S.

66. We have to prove that $$\sin A\sin(60^\circ + A)\sin(A + 120^\circ) = \sin 3A$$

We have proved in problem 32 that $$\sin A\sin(60^\circ - A)\sin(60^\circ + A) = \frac{1}{4}\sin 3A$$

Thus, we can prove what is required.

67. We have to prove that $$\cot A + \cot(60^\circ + A) + \cot(120^\circ + A) = 3\cot 3A$$

L.H.S. $$= \cot A + \cot(60^\circ + A) + \cot(180^\circ - (60^\circ - A))$$

$$= \cot A + \cot(60^\circ + A) - \cot(60^\circ - A)$$

This we have proved in problem 34.

68. We have to prove that $$\cos 5A = 16\cos^5A - 20\cos^3A + 5\cos A$$

L.H.S. $$\cos(2A + 3A) = \cos 2A\cos 3A - \sin2A\sin3A$$

$$= (2\cos^2A - 1)(4\cos^3A - 3\cos A) - 2\sin A\cos A(3\sin A - 4\sin^3A)$$

$$= 8\cos^5A - 10\cos^3A + 3\cos A - 2\cos A\sin^2A[3 - 4(1 - \cos^2A)]$$

$$= 8\cos^5A - 10\cos^3A + 3\cos A - 2\cos A(1 - \sin^2A)[4\cos^2A - 1]$$

$$= 16\cos^5A - 20\cos^3A + 5\cos A =$$ R.H.S.

69. We have to prove that $$\sin 5A = 5\sin A - 20\sin^3A + 16\sin^5A$$

L.H.S. $$= \sin5A = \sin(2A + 3A) = \sin2A\cos3A + \sin3A\cos2A$$

$$= 2\sin A\cos A(4\cos^3A - 3\cos A) + (3\sin A - 4\sin^3A)(1 - 2\sin^2A)$$

$$= 2\sin A(1 - \sin^2A)(4\cos^2A - 3) + (3\sin A - 4\sin^3A)(1 - 2\sin^2A)$$

$$= 2(\sin A - \sin^3A)(1 - 4\sin^2A) + (3\sin A - 4\sin^3A)(1 - 2\sin^2A)$$

$$= 5\sin A - 20\sin^3A + 16\sin^5A =$$ R.H.S.

70. We have to prove that $$\cos 4A - \cos 4B = 8(\cos A - \cos B)(\cos A + \cos B)(\cos A - \sin B)(\cos A + \sin B)$$

R.H.S. $$= 2(2\cos^2A - 2\cos^2B)(2\cos^2A - 2\sin^2B)$$

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

$$= 2(\cos^22A - \cos^22B) = \cos4A - \cos4B =$$ L.H.S.

71. We have to prove that $$\tan 4A = \frac{4\tan A - 4\tan^3A}{1 - 6\tan^2A + \tan^4A}$$

L.H.S. $$= \tan 4A = \tan(2A + 2A) = \frac{2\tan2A}{1 - \tan^22A}$$

$$= \frac{2.\frac{2\tan A}{1 - \tan^2A}}{1 - \left(\frac{2\tan A}{1 - \tan^2A}\right)^2}$$

Solving this yields R.H.S.

72. Given $$2\tan A = 3\tan B,$$ we have to prove that $$\tan (A- B) = \frac{\sin 2B}{5 - \cos 2B}$$

$$\tan A = \frac{3}{2}\tan B$$

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

$$= \frac{\tan B}{2 + 3\tan^2B} = \frac{\sin B\cos B}{2\cos^2B + 3\sin^2B}$$

$$= \frac{\sin B\cos B}{1 + \cos 2B + 3.\frac{1}{2}(1 - \cos 2B)}$$

$$= \frac{\sin 2B}{5 - \cos 2B} =$$ R.H.S.

73. Given $$\sin A + \sin B = x$$ and $$\cos A + \cos B = y,$$ we have to show that $$\sin(A + B) = \frac{2xy}{x^2 + y^2}$$

$$2xy = 2(\sin A + \sin B)(\cos A + \cos B)$$

$$= 2(\sin A\cos A + \sin B\cos B + \sin A\cos B + \cos A\sin B)$$

$$= \sin2A + \sin 2B + 2\sin(A + B)$$

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

$$x^2 + y^2 = (\sin A + \sin B)^2 + (\cos A + \cos B)^2$$

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

$$\therefore \frac{2xy}{x^2 + y^2} = \sin(A + B)$$

74. Given $$A= \frac{\pi}{2^n + 1},$$ we have to prove that $$\cos A.\cos 2A. \cos2^2A.\ldots.\cos2^{n - 1}A = \frac{1}{2^n}$$

L.H.S. $$= \cos A.\cos 2A. \cos2^2A.\ldots.\cos2^{n - 1}A$$

$$= \frac{1}{2\sin A}(2\sin A\cos A).\cos 2A. \cos2^2A.\ldots.\cos2^{n - 1}A$$

$$= \frac{1}{2\sin A}\sin 2A.\cos 2A.\cos2^2A.\ldots.\cos2^{n - 1}A$$

$$= \frac{1}{2^2\sin A}(2\sin 2A\cos 2A)\cos2^2A.\ldots.\cos2^{n - 1}A$$

Proceeding similarly

$$= \frac{1}{2^n\sin A}\sin2^n A = \frac{1}{2^n\sin A}\sin(\pi - A) = \frac{1}{2^n} =$$ R.H.S.

75. Given $$\tan A = \frac{y}{x},$$ we have to prove that $$x\cos 2A + y\sin 2A = x$$

$$\because \tan A = \frac{y}{x} \therefore \sin A = \frac{y}{\sqrt{x^2 + y^2}}, \cos A = \frac{x}{\sqrt{x^2 + y^2}}$$

$$\therefore x\cos 2A + y\sin 2A = x(\cos^2A - \sin^2A) + 2y\sin A\cos A = x\left(\frac{x^2 - y^2}{x^2 + y^2}\right) + 2\frac{x^2y}{x^2 + y^2}$$

$$= x =$$ R.H.S.

76. Given $$\tan^2A = 1 + 2\tan^2B,$$ we have to prove that $$\cos 2B = 1 + 2\cos 2A$$

$$1 + 2\cos 2A = 1 + 2.\frac{1 - \tan^2A}{1 + \tan^2A} = \frac{3 - \tan^2A}{1 + \tan^2A}$$

$$= \frac{3 - 1 - 2\tan^2B}{2 + 2\tan^2B} = \frac{1 - \tan^2B}{1 + \tan^2B} = \cos 2B =$$ L.H.S.

77. Given $$\cos 2A = \frac{3\cos 2B - 1}{3 - \cos 2B},$$ we have to prove that $$\tan A = \sqrt{2}\tan B$$

$$\cos 2A = \frac{3\cos 2B - 1}{3 - \cos 2B}$$

$$\Rightarrow \frac{1 - \tan^2A}{1 + \tan^B} = \frac{3 - 3\tan^2B - 1 - \tan^2B}{3 + 3\tan^2B - 1 + \tan^2B}$$

$$= \frac{1 - 2\tan^2B}{1 + 2\tan^2B}$$

$$\therefore \tan^2A = 2\tan^2B \Rightarrow \tan A = \sqrt{2}\tan B$$

78. Given $$\tan B = 3\tan A,$$ we have to prove that $$\tan(A + B) = \frac{2\sin 2B}{1 + \cos 2B}$$

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

$$= \frac{\frac{4}{3}\tan B}{1 - \frac{\tan^2B}{3}} = \frac{4\tan B}{3 - \tan^2B}$$

$$= \frac{4\sin B\cos B}{3\cos^2B - \sin^2B} = \frac{2\sin2B}{2\cos^2B + \cos2B}$$

$$= \frac{2\sin2B}{1 + \cos2B} =$$ R.H.S.

79. Given $$x\sin A = y\cos A,$$ we have to prove that $$\frac{x}{\sec 2A} + \frac{y}{\cosec 2A} = x$$

Given $$\tan A = \frac{y}{x} \therefore \sin A = \frac{y}{\sqrt{x^2 + y^2}} \& \cos A = \frac{x}{\sqrt{x^2 + y^2}}$$

L.H.S. $$= \frac{x}{\sec 2A} + \frac{y}{\cosec 2A}$$

$$= x\cos2A + y\sin2A = x(\cos^2A - \sin^2A) + 2y\sin A\cos A$$

$$x\frac{x^2 - y^2}{x^2 + y^2} + \frac{2xy^2}{x^2 + y^2}$$

$$= x$$

80. Given $$\tan A = \sec 2B,$$ we have to prove that $$\sin 2A = \frac{1 - \tan^4B}{1 + \tan^4B}$$

$$\tan A = \frac{1}{\cos 2B} = \frac{1 + \tan^2B}{1 - \tan^2B}$$

$$\therefore \sin A = \frac{1 + \tan^2B}{\sqrt{2 + 2\tan^4B}}$$

and $$\cos A = \frac{1 - \tan^2B}{\sqrt{2 + 2\tan^4B}}$$

L.H.S. $$\sin 2A = 2\sin A\cos A = \frac{1 - \tan^4B}{1 + \tan^4B} =$$ R.H.S.

81. Given $$A = \frac{\pi}{3},$$ we have to prove that $$\cos A.\cos 2A. \cos 3A.\cos 4A.\cos 5A.\cos 6A = -\frac{1}{16}$$

L.H.S. $$= \frac{1}{8}2\cos A.\cos 6A.2\cos 2A.\cos 5A.2\cos 3A\cos 4A$$

$$= \frac{1}{8}\left(\cos \frac{7A}{2} + \cos \frac{5A}{2}\right)\left(\cos \frac{7A}{2} + \cos \frac{3A}{2}\right)\left(\cos \frac{7A}{2} + \cos \frac{A}{2}\right)$$

$$= \frac{1}{8}\left[\cos\left(\pi + \frac{\pi}{6}\right) + \cos \left(\pi - \frac{\pi}{6}\right)\right]\left[\cos\left(\pi + \frac{\pi}{6}\right) + \cos \left(\frac{\pi}{2}\right)\right] + \left[\cos\left(\pi + \frac{\pi}{6}\right) + \cos \frac{\pi}{6}\right]$$

$$= -\frac{1}{16}$$

82. Given $$A = \frac{\pi}{15},$$ we have to prove that $$\cos2A.\cos4A.\cos8A.\cos14A = \frac{1}{16}$$

$$\cos 14A = \cos \frac{14\pi}{15} = \cos \left(2\pi - \frac{16\pi}{15}\right) = \cos 16A$$

L.H.S. $$= \cos2A.\cos4A.\cos8A.\cos16A = \frac{1}{2\sin2A}.2\sin2A.\cos2A.\cos4A.\cos8A.\cos16A$$

$$= \frac{1}{2\sin2A}\sin4A.\cos4A.\cos8A.\cos16A = \frac{1}{2^2\sin 2A}\sin8A.\cos8A.\cos16A$$

$$= \frac{1}{2^4\sin 2A}\sin32A = \frac{1}{16\sin2A}\sin(2\pi + 2A) = \frac{1}{16} =$$ R.H.S.

83. Given $$\tan A\tan B = \sqrt{\frac{a - b}{a + b}},$$ we have to prove that $$(a - b\cos2A)(a - b\cos2B) = a^2 - b^2$$

L.H.S. $$= (a - b\cos2A)(a - b\cos2B) = \left[a -b\frac{1 - \tan^2A}{1 + \tan^2A}\right]\left[a - b\frac{1 - \tan^2B}{1 + \tan^2B}\right]$$

$$= \left[a -b\frac{1 - \tan^2A}{1 + \tan^2A}\right]\left[a - b\frac{1 - \frac{a - b}{(a + b)\tan^2A}}{1 + \frac{a - b}{(a + b)\tan^2A}}\right]$$

Solving this yields $$\frac{a^2 - b^2}{}$$

84. Given $$\sin A = \frac{1}{2}$$ and $$\sin B = \frac{1}{3},$$ we have to find the value of $$\sin(A + B)$$ and $$\sin(2A + 2B)$$

$$\cos A = \frac{\sqrt{3}}{2}$$ and $$\cos B = \frac{\sqrt{8}}{3}$$

$$\sin(A + B) = \sin A\cos B + \cos A\sin B = \frac{\sqrt{8}}{6} + \frac{\sqrt{3}}{6} = \frac{\sqrt{8} + \sqrt{3}}{6}$$

$$\sin(2A + 2B) = \sin 2A\cos 2B + \cos 2A\sin 2B$$

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

Substituting the values we obtain the desired result.

85 and 86 have been left as exercises.

1. $$\cos A = \frac{3}{10} = \frac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}} = \frac{3}{10}$$

Let $$x = \tan \frac{A}{2},$$ then $$\frac{1 - x^2}{1 + x^2} = \frac{3}{10}$$

$$x = \pm \sqrt{\frac{7}{13}}$$

The reason for two values is that $$\cos A$$ may lie in first or fourth quadrant. If it is in first quadrant then $$\tan \frac{A}{2}$$ will be positive and if it is in fourth quadrant then $$\tan \frac{A}{2}$$ will be negative.

2. Given $$\sin A + \sin B = x$$ and $$\cos A + \cos B = y,$$ we have to find the value of $$\tan \frac{A - B}{2}$$

$$\tan \frac{A - B}{2} = \frac{\tan \frac{A}{2} - \tan \frac{B}{2}}{1 + \tan\frac{A}{2}\tan\frac{B}{2}}$$

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

Also, $$\tan(A - B) = \frac{2\tan\frac{A - B}{2}}{1 - \tan^2\frac{A - B}{2}}$$

Let $$\tan\frac{A - B}{2} = a,$$ then $$\tan(A - B) = \frac{2a}{1 + a^2}$$

$$x^2 + y^2 = 2 + 2\sin A\sin B + 2\cos A\cos B$$

Solving this yields $$\tan\frac{A - B}{2} = \sqrt{\frac{4 - x^2 - y^2}{x^2 + y^2}}$$

3. We have to prove that $$(\cos A + \cos B)^2 + (\sin A - \sin B)^2 = 4\cos^2 \frac{A + B}{2}$$

L.H.S. $$= \cos^2A + \cos^2B + 2\cos A\cos B + \sin^2A + \sin^2B - 2\sin A\sin B$$

$$= 2 + 2\cos(A + B) = 4\cos^2\frac{A + B}{2} =$$ R.H.S.

4. We have to prove that $$(\cos A + \cos B)^2 + (\sin A + \sin B)^2 = 4\cos^2 \frac{A - B}{2}$$

L.H.S. $$= \cos^2A + \cos^2B + 2\cos A\cos B + \sin^2A + \sin^2B + 2\sin A\sin B$$

$$= 2 + 2\cos(A - B) = 4\cos^2\frac{A - B}{2} =$$ R.H.S.

5. We hve to prove that $$(\cos A - \cos B)^2 + (\sin A - \sin B)^2 = 4\sin^2 \frac{A - B}{2}$$

L.H.S. $$= \cos^2A + \cos^2B - 2\cos A\cos B + \sin^2A + \sin^2B + 2\sin A\sin B$$

$$= 2 - 2\cos(A - B) = 4\sin^2\frac{A - B}{2} =$$ R.H.S.

6. We have to prove that $$\sin^2\left(\frac{\pi}{8} + \frac{A}{2}\right) - \sin^2\left(\frac{\pi}{8} -\frac{A}{2}\right) = \frac{1}{\sqrt{2}}\sin A$$

L.H.S. $$= \frac{1 - \cos\left(\frac{\pi}{4} + A\right)}{2} - \frac{1 - \cos\left(\frac{\pi}{4} - A\right)}{2}$$

$$= \frac{\cos\left(\frac{\pi}{4} - A\right) - \cos\left(\frac{\pi}{4} + A\right)}{2}$$

$$= \frac{2\sin\frac{\pi}{4}\sin A}{2} = \frac{1}{\sqrt{2}}\sin A =$$ R.H.S.

7. We have to prove that $$(\tan 4A + \tan 2A)(1 - \tan^23A\tan^2A) = 2\tan 3A\sec^2A$$

L.H.S. $$= (\tan 4A + \tan 2A)(1 + \tan 3A\tan A)(1 - \tan 3A\tan A)$$

$$= \left(\frac{\sin 4A}{\cos 4A} + \frac{\sin 2A}{\cos 2A}\right)\left(\frac{cos 3A\cos A + \sin 3A\sin A}{\cos 3A\cos A}\right)\left(\frac{cos 3A\cos A - \sin 3A\sin A}{\cos 3A - \cos A}\right)$$

$$= \frac{\sin 6A}{\cos 4A\cos 2A}.\frac{\cos 4A}{\cos 3A\cos A}\frac{\cos 2A}{\cos 3A\cos A}$$

$$= \frac{2\sin3A\cos3A}{\cos^23A\cos^2A} = 2\tan3A\sec^2A =$$ R.H.S.

8. We have to prove that $$\left(1 + \tan \frac{A}{2} - \sec\frac{A}{2}\right)\left(1 + \tan \frac{A}{2} + \sec\frac{A}{2}\right) = \sin A\sec^2\frac{A}{2}$$

L.H.S. $$= \left(1 + \tan \frac{A}{2} - \sec\frac{A}{2}\right)\left(1 + \tan \frac{A}{2} + \sec\frac{A}{2}\right)$$

$$= \left(1 + \tan\frac{A}{2}\right)^2 - sec^2\frac{A}{2} = 2\tan\frac{A}{2}$$

$$= \frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{\cos^2\frac{A}{2}} = \sin A\sec^2\frac{A}{2} =$$ R.H.S.

9. We have to prove that $$\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \frac{A}{2}$$

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

$$= \frac{2\sin^2\frac{A}{2} + 2\sin\frac{A}{2}\cos\frac{A}{2}}{2\cos^2\frac{A}{2} + 2\sin\frac{A}{2}\cos\frac{A}{2}}$$

$$= \frac{\sin\frac{A}{2}\left(\sin\frac{A}{2} + \cos\frac{A}{2}\right)}{\cos\frac{A}{2}\left(\sin\frac{A}{2} + \cos\frac{A}{2}\right)}$$

$$= \tan\frac{A}{2} =$$ R.H.S.

10. We have to prove that $$\frac{1 - \tan \frac{A}{2}}{1 + \tan \frac{A}{2}} = \frac{1 + \sin A}{\cos A} = \tan \left(\frac{\pi}{4} + \frac{A}{2}\right)$$

$$\frac{1 + \sin A}{\cos A} = \frac{\sin^2\frac{A}{2} + \cos^2\frac{A}{2} + 2\sin\frac{A}{2}\cos \frac{A}{2}}{\cos^2\frac{A}{2} - \sin^2\frac{A}{2}}$$

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

Dividing numerator and denominator by $$\cos \frac{A}{2},$$ we get

$$= \frac{1 + \tan \frac{A}{2}}{1 - \tan \frac{A}{2}}$$

11. We have to prove that $$\cos^4\frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4\frac{5\pi}{8} + \cos^4\frac{7\pi}{8}= \frac{3}{2}$$

$$\cos^4\frac{\pi}{8} = \left(\cos^2\frac{\pi}{8}\right)^2 = \left(\frac{1 + \cos\frac{\pi}{4}}{2}\right)^2$$

$$= \left(\frac{1 + \frac{1}{\sqrt{2}}}{2}\right)^2 = \frac{3}{8} + \frac{\sqrt{2}}{4}$$

Similalry, $$\cos^4\frac{3\pi}{8} = \frac{3}{8} - \frac{\sqrt{2}}{4}$$

$$\cos\frac{5\pi}{8} = \cos\left(\pi - \frac{3\pi}{8}\right) = -\cos\frac{3\pi}{8}$$

$$\cos\frac{7\pi}{8} = -\cos\frac{\pi}{8}$$

Thus, $$\cos^4\frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4\frac{5\pi}{8} + \cos^4\frac{7\pi}{8}= \frac{3}{2}$$

12. We have to prove that $$\frac{2\sin A - \sin2A}{2\sin A + \sin 2A} = \tan^2\frac{A}{2}$$

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

$$= \frac{2\sin^2\frac{A}{2}}{2\cos^2\frac{A}{2}} = \tan^2\frac{A}{2} =$$ R.H.S.

13. We have to prove that $$\cot \frac{A}{2} - \tan \frac{A}{2} = 2\cot A$$

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

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

$$= \frac{2\cos A}{\sin A} = 2\cot A =$$ R.H.S.

14. We have to prove that $$\frac{1 + \sin A}{1 - \sin A} = \tan^2\left(\frac{\pi}{4} + \frac{A}{2}\right)$$

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

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

Dividing both numerator and denominator by $$\cos \frac{A}{2},$$ we get

$$= \frac{1 + \tan\frac{A}{2}}{1 - \tan \frac{A}{2}} = \frac{\tan\frac{\pi}{4} + \tan \frac{A}{2}}{1 - \tan\frac{\pi}{4}\tan\frac{A}{2}}$$

$$= \tan\left(\frac{\pi}{4} + \frac{A}{2}\right) =$$ R.H.S.

15. We have to prove that $$\sec A + \tan A = \tan\left(\frac{\pi}{4} + \frac{A}{2}\right)$$

L.H.S. $$= \frac{1 + \sin A}{\cos A} = \frac{\left(\cos\frac{A}{2} + \sin\frac{A}{2}\right)^2}{\cos^2\frac{A}{2} - \sin^2\frac{A}{2}}$$

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

Now proceeding like previous problem

$$= \tan\left(\frac{\pi}{4} + \frac{A}{2}\right) =$$ R.H.S.

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

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

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

$$= \frac{2\sin\frac{A}{2}\cos\frac{B}{2}}{2\cos\frac{A}{2}\cos\frac{B}{2}}$$

$$= \tan \frac{A}{2}\tan \frac{B}{2} =$$ R.H.S.

17. We have to prove that $$\tan \left(\frac{\pi}{4} - \frac{A}{2}\right) = \sec A - \tan A = \sqrt{\frac{1 - \sin A}{1 + \sin A}}$$

L.H.S. $$= \tan \left(\frac{\pi}{4} - \frac{A}{2}\right) = \frac{1 - \tan \frac{A}{2}}{1 + \tan \frac{A}{2}}$$

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

Multiplying both numerator and denominator by $$\cos\frac{A}{2} + \sin\frac{A}{2}$$

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

Also, $$\frac{\cos A}{1 + \sin A} = \frac{\cos A(1 - \sin A)}{1 - \sin^2A} = \sec A - \tan A$$

18. We have to prove that $$\cosec\left(\frac{\pi}{4} + \frac{A}{2}\right)\cosec \left(\frac{\pi}{4} - \frac{A}{2}\right) = 2\sec A$$

L.H.S. $$= \frac{1}{\sin\left(\frac{\pi}{4} + \frac{A}{2}\right)}.\frac{1}{\sin\left(\frac{\pi}{4} - \frac{A}{2}\right)}$$

$$= \frac{2}{\cos A - \cos \frac{\pi}{2}} = 2\sec A =$$ R.H.S.

19. We have to prove that $$\cos^2\frac{\pi}{8} + \cos^2\frac{3\pi}{8} + \cos^2\frac{5\pi}{8} + \cos^2\frac{7\pi}{8} = 2$$

$$\cos^2\frac{\pi}{8} = \frac{1 + \cos \frac{\pi}{4}}{2} = \frac{1 + \sqrt{2}}{2\sqrt{2}}$$

$$\cos^2\frac{3\pi}{8} = \frac{1 + \cos \frac{3\pi}{4}}{2} = \frac{\sqrt{2} - 1}{2\sqrt{2}}$$

$$\cos^2\frac{5\pi}{8} = \cos^2\frac{3\pi}{8}$$

$$\cos^2\frac{7\pi}{8} = \cos^2\frac{\pi}{8}$$

L.H.S. $$= 2\left(\frac{1 + \sqrt{2}}{2\sqrt{2}} + \frac{\sqrt{2} - 1}{2\sqrt{2}}\right)$$

$$= 2 =$$ R.H.S.

20. This problem is similar to previous problem and can be solved in a likewise manner.

21. We have to prove that $$\left(1 + \cos \frac{\pi}{8}\right)\left(1 + \cos\frac{3\pi}{8}\right)\left(1 + \cos\frac{5\pi}{8}\right)\left(1 + \cos \frac{7\pi}{8}\right) = \frac{1}{8}$$

$$\cos\frac{7\pi}{8} = \cos\left(\pi - \frac{\pi}{8}\right) = -\cos\frac{\pi}{8}$$

$$\cos\frac{5\pi}{8} = \cos\left(\pi - \frac{3\pi}{8}\right) = -\cos\frac{3\pi}{8}$$

L.H.S. $$= \left(1 - \cos^2\frac{\pi}{8}\right)\left(1 - \cos^2\frac{3\pi}{8}\right)$$

$$= \sin^2\frac{\pi}{8}\sin^2\frac{3\pi}{8}$$

$$= \frac{1 - 2\cos\frac{\pi}{4}}{2}.\frac{1 - 2\cos\frac{3\pi}{4}}{2}$$

Substituting values from 105 we get desired result.

22. We have to find the value of $$\sin \frac{23\pi}{24}$$

$$\sin \left(\pi - \frac{\pi}{24}\right) = \sin\frac{15^\circ}{2}$$

$$\sin^2A = \frac{1}{2}(1 - \cos 2A) = \frac{1}{2}(1 - \cos15^\circ)$$

$$= \frac{1}{2}\left(1 - \frac{\sqrt{3} + 1}{2\sqrt{2}}\right)$$

$$\therefore \sin A = \frac{1}{4}\sqrt{8 - 2\sqrt{6} - 2\sqrt{2}}$$

23. Given $$A = 112^\circ30'\therefore 2A = 225^\circ$$

$$\cos 2A = \cos(180^\circ + 45^\circ) = -\frac{1}{\sqrt{2}}$$

$$|\sin A| = \sqrt{\frac{1 - \left(-\frac{1}{\sqrt{2}}\right)}{2}} = \sqrt{\frac{2 + \sqrt{2}}{2}}$$

$$\because$$ A lies in 2nd quadrant $$\therefore \sin A$$ will be positive and $$\cos A$$ will be negative.

$$|\cos A| = -\frac{\sqrt{2 - \sqrt{2}}}{2}$$

24. We have to prove that $$\sin^224^\circ - \sin^26^\circ = \frac{1}{8}(\sqrt{5} - 1)$$

L.H.S. $$= \sin(24^\circ + 6^\circ)\sin(24^\circ - 6^\circ) = \frac{1}{2}.\frac{\sqrt{5} - 1}{4}$$

$$= \frac{1}{8}(\sqrt{5} - 1) =$$ R.H.S.

25. We have to prove that $$\tan6^\circ.\tan42^\circ.\tan66^\circ.\tan78^\circ = 1$$

L.H.S. $$= \frac{\sin66^\circ6^\circ}{\cos66^\circ\cos6^\circ}.\frac{\sin78^\circ\sin42^\circ}{\cos78^\circ\cos42^\circ}$$

$$= \frac{\cos60^\circ - \cos72^\circ}{\cos60^\circ + \cos72^\circ}.\frac{\cos36^\circ - \cos120^\circ}{\cos36^\circ + \cos120^\circ}$$

$$= \frac{1 - 2\sin18^\circ}{1 + 2\sin18^\circ}.\frac{2\cos36^\circ + 1}{2\cos36^\circ - 1}$$

$$= \frac{1 - 2\left(\frac{\sqrt{5} - 1}{4}\right)}{1 + 2\left(\frac{\sqrt{5} - 1}{4}\right)}.\frac{2.\left(\frac{\sqrt{5} + 1}{4}\right) + 1}{2.\left(\frac{\sqrt{5} + 1}{4}\right) - 1}$$

$$= 1 =$$ R.H.S.

26. We have to prove that $$\sin47^\circ + \sin61^\circ - \sin 11^\circ - \sin25^\circ = \cos 7^\circ$$

L.H.S. $$= 2\sin54^\circ\cos7^\circ - 2\sin18^\circ\cos7^\circ$$

$$= 2\cos7^\circ.2\cos36^\circ.\sin18^\circ = 2\cos7^\circ.2\frac{\sqrt{5} + 1}{4}.\frac{\sqrt{5} - 1}{4}$$

$$= \cos7^\circ$$

27. We have to prove that $$\sin 12^\circ\sin48^\circ\sin54^\circ = \frac{1}{8}$$

L.H.S. $$= \frac{1}{2}.2\sin48^\circ\sin12^\circ.\sin54^\circ$$

$$= \frac{1}{2}(\cos 36^\circ - \cos60^\circ).\cos36^\circ$$

$$= \frac{1}{2}\left(\frac{\sqrt{5} + 1}{4} - \frac{1}{2}\right).\frac{\sqrt{5} + 1}{4}$$

$$= \frac{1}{8} =$$ R.H.S.

28. We have to prove that $$\cot 142\frac{1}{2}^\circ = \sqrt{2} + \sqrt{3} - 2 - \sqrt{6}$$

L.H.S. $$\cos 142\frac{1}{2}^\circ = \cot\left(180^\circ - 37\frac{1}{2}^\circ\right) = -\cot37\frac{1}{2}^\circ$$

We know that $$\tan 15^\circ = \cot 75^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$$

$$\therefore -\cot37\frac{1}{2}^\circ = \sqrt{2} + \sqrt{3} - 2 - \sqrt{6} =$$ R.H.S.

29. We have to prove that $$\sin^248^\circ - \cos^212^\circ = -\frac{\sqrt{5} + 1}{8}$$

L.H.S. $$= \frac{1}{2}\left(2\sin^248^\circ - 2\cos^212^\circ\right)$$

$$= \frac{1}{2}\left(1 - \cos96^\circ \right - 1 - \cos24^\circ)$$

$$= -\frac{1}{2}\left(2\cos60^\circ\cos36^\circ\right)$$

$$= -\frac{1}{2}.\frac{\sqrt{5} + 1}{4} = -\frac{\sqrt{5} + 1}{8} =$$ R.H.S.

30. We have to prove that $$4(\sin 24^\circ + \cos6^\circ) = \sqrt{3} + \sqrt{15}$$

L.H.S. $$= 4(\sin24^\circ + \sin84^\circ) = 8\sin54^\circ\cos30^\circ = 4\sqrt{3}\sin54^\circ$$

$$= 4\sqrt{3}(3\sin18^\circ - 4\sin^318^\circ)$$

We know that $$\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$$

$$\therefore 4\sqrt{3}(3\sin18^\circ - 4\sin^318^\circ) = \sqrt{3} + \sqrt{15} =$$ R.H.S.

31. We have to prove that $$\cot6^\circ\cot42^\circ\cot66^\circ\cot78^\circ = 1$$

L.H.S. $$= \frac{1}{\tan6^\circ\tan42^\circ\tan66^\circ\tan78^\circ}$$

We know that $$\tan(60^\circ - x)\tan x\tan(60^\circ + x) = \tan 3x$$

Putting $$x=18^\circ,$$ we get

$$\tan42^\circ\tan18^\circ\tan78^\circ = \tan54^\circ$$

Putting $$x=6^\circ,$$ we get

$$\tan54^\circ\tan6^\circ\tan66^\circ = \tan18^\circ$$

From these two, we derive that

$$\tan6^\circ\tan42^\circ\tan66^\circ\tan78^\circ = 1$$

32. We have to prove that $$\tan12^\circ\tan24^\circ\tan48^\circ\tan84^\circ = 1$$

We know that $$\tan(60^\circ - x)\tan x\tan(60^\circ + x) = \tan 3x$$

Putting $$x= 12^\circ,$$ we get

$$\tan48^\circ\tan 12^\circ\tan72^\circ = \tan 36^\circ$$

Putthing $$x = 24^\circ,$$ we get

$$\tan36^\circ\tan24^\circ\tan84^\circ = \tan72^\circ$$

From these two, we derive that

$$\tan12^\circ\tan24^\circ\tan48^\circ\tan84^\circ = 1$$

33. We have to prove that $$\sin6^\circ\sin42^\circ\sin66^\circ\sin78^\circ = \frac{1}{16}$$

L.H.S. $$= \sin6^\circ\sin66^\circ\sin42^\circ\sin78^\circ$$

$$= \frac{1}{4}(\cos60^\circ - \cos72^\circ)(\cos 36^\circ - \cos120^\circ)$$

$$= \frac{1}{4}\left(\frac{1}{2} - \cos72^\circ\right)\left(\cos36^\circ + \frac{1}{2}\right)$$

$$= \frac{1}{16}(1 - 2\cos72^\circ)(2\cos36^\circ + 1)$$

$$= \frac{1}{16}[1 + 2\cos36^\circ - 2\cos72^\circ - 4\cos36^\circ\cos72^\circ]$$

$$= \frac{1}{16} + \frac{1}{8}[\cos36^\circ - \cos72^\circ - \cos 108^\circ - \cos36^\circ]$$

$$= \frac{1}{16} + \frac{1}{8}[\cos72^\circ + \circ108^\circ]$$

$$= \frac{1}{16} + \frac{1}{8}[\cos72^\circ + \cos(180^\circ - 72^\circ)]$$

$$= \frac{1}{16}$$

34. We have to prove that $$\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}$$

L.H.S. $$= \sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\left(\pi - \frac{2\pi}{5}\right)\sin\left(\pi - \frac{\pi}{5}\right)$$

$$= \sin^2\frac{\pi}{5}\sin^2\frac{2\pi}{5} = \sin^218^\circ\sin^236^\circ = \left(\frac{\sqrt{5} - 1}{4}\right)^2\left(\frac{1}{4}\sqrt{10 - 2\sqrt{5}}\right)^2$$

$$= \frac{5}{16} =$$ R.H.S.

35. We have to prove that $$\cos36^\circ\cos72^\circ\cos108^\circ\cos144^\circ = \frac{1}{16}$$

L.H.S. $$= \cos36^\circ\cos72^\circ\cos(180^\circ - 72^\circ)\cos(180^\circ - 36^\circ)$$

$$= \cos^236^\circ\cos^272^\circ$$

$$\cos 36^\circ = \frac{\sqrt{5} + 1}{4}, \cos72^\circ = 2\cos^236^\circ - 1$$

Thus, $$\cos^236^\circ\cos^272^\circ = \frac{1}{16}$$

36. We have to prove that $$\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15} = \frac{1}{2^7}$$

L.H.S. $$= \frac{1}{2\sin\frac{\pi}{15}}2\sin\frac{\pi}{15}\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}$$

$$= \frac{1}{2\sin\frac{\pi}{15}}\sin\frac{2\pi}{15}\cos\frac{2\pi}{15}\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}$$

$$= \frac{1}{2^2\sin\frac{\pi}{15}}\sin\frac{4\pi}{15}\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}$$

$$= \frac{1}{2^3\sin\frac{\pi}{15}}\sin\frac{8\pi}{15}\cos\frac{3\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}$$

Now, $$\sin\frac{8\pi}{15} = \sin\left(\pi - \frac{7\pi}{15}\right) = \sin\frac{7\pi}{15},$$ therefore

$$= \frac{1}{2^4\sin\frac{\pi}{15}}2\sin\frac{7\pi}{15}\cos\frac{3\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}$$

$$= \frac{1}{2^4\sin\frac{\pi}{15}}\sin\frac{14\pi}{15}\cos\frac{3\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}$$

Now $$\sin\frac{14\pi}{15} = \sin\left(\pi - \frac{\pi}{15}\right) = \sin\frac{\pi}{15},$$ therefore

$$= \frac{1}{2^4}\cos\frac{3\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}$$

$$= \frac{1}{2^5\sin\frac{3\pi}{15}}2\sin\frac{3\pi}{15}\cos\frac{3\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}$$

$$= \frac{1}{2^6\sin\frac{3\pi}{15}}2\sin\frac{6\pi}{15}\cos\frac{6\pi}{15}\cos\frac{\pi}{3}$$

$$= \frac{1}{2^6\sin\frac{3\pi}{15}}\sin\frac{12\pi}{15}\cos\frac{\pi}{3}$$

Similarly $$\sin\frac{12\pi}{15} = \sin \frac{3\pi}{15}$$

$$= \frac{1}{2^6}\cos\frac{\pi}{3} = \frac{1}{2^7} =$$ R.H.S.

37. We have to prove that $$\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65} = \frac{1}{64}$$

$$= \frac{1}{2\sin\frac{\pi}{65}}2\sin\frac{\pi}{65}\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$$

$$= \frac{1}{2\sin\frac{\pi}{65}}\sin\frac{2\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$$

$$= \frac{1}{2^2\sin\frac{\pi}{65}}2\sin\frac{2\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$$

$$= \frac{1}{2^2\sin\frac{\pi}{65}}\sin\frac{4\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$$

$$= \frac{1}{2^3\sin\frac{\pi}{65}}2\sin\frac{4\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$$

Proceeding similalry we find that above is equal to

$$\frac{1}{2^7\sin\frac{\pi}{65}}\sin\frac{64\pi}{65}$$

However, $$\sin\frac{64\pi}{65} = \sin\frac{\pi}{65},$$ therefore

$$\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65} = \frac{1}{64}$$

38. Given, $$\tan \frac{A}{2} = \sqrt{\frac{a - b}{a + b}}\tan \frac{B}{2}$$

Now, $$\cos A = \frac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}} = \frac{1 - \frac{a - b}{a + b}\tan^2\frac{B}{2}}{1 + \frac{a - b}{a + b}\tan^2\frac{B}{2}}$$

$$= \frac{(a + b)\cos^2\frac{B}{2} - (a - b)\sin^2\frac{B}{2}}{(a + b)\cos^2\frac{B}{2} + (a - b)\sin^2\frac{B}{2}}$$

$$= \frac{a\cos B + b}{a + b\cos B}$$

39. This problem is similar to previous problem with $$a = 1, e = b$$ and has been left as an exercise.

40. Given $$\sin A + \sin B = a$$ and $$\cos A + \cos B = b,$$ we have to prove that $$\sin(A + B) = \frac{2ab}{a^2 + b^2}$$

$$2ab = 2(\sin A + \sin B)(\cos A + \cos B) = 2\sin A\cos A + 2\sin A\cos B + 2\sin B\cos A + 2\sin B\cos B = \sin 2A + \sin 2B + 2\sin(A + B)$$

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

$$a^2 + b^2 = \sin^2A + \sin^2B + 2\sin A\sin B + \cos^2A + \cos^2B + 2\cos A\cos B$$

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

$$\therefore \sin(A + B) = \frac{2ab}{a^2 + b^2}$$

41. Given $$\sin A + \sin B = a$$ and $$\cos A + \cos B = b,$$ we have to prove that $$\cos(A - B) = \frac{1}{2}(a^2 + b^2 - 2)$$

From previous problem, $$2\cos(B - A) = a^2 + b^2 - 2 \Rightarrow \cos(A - B) = \frac{1}{2}(a^2 + b^2 - 2)$$

42. Let us solve these one by one.

1. Given $$A$$ and $$B$$ be two different roots of equation $$a\cos\theta + b\sin\theta = c$$

$$a\cos A + b\sin A = c$$ and $$a\cos B + b\sin B = c$$

$$\Rightarrow a(\cos A - \cos B) + b(\sin A - \sin B) = 0$$

$$b(\sin A - \sin B) = a(\cos A - \cos B)$$

$$b.2\cos\frac{A + B}{2}\sin\frac{A - B}{2} = a.2\sin\frac{A + B}{2}\sin \frac{A - B}{2}$$

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

$$\tan{A + B} = \frac{2\tan\frac{A + B}{2}}{1 - \tan^2\frac{A + B}{2}} = \frac{2ab}{a^2 + b^2}$$

2. We have $$\tan(A + B) = \frac{2ab}{a^2 + b^2}$$

$$\therefore \cos(A + B) = \frac{a^2 - b^2}{a^2 + b^2}$$

43. Given $$\cos A + \cos B = \frac{1}{3}$$ and $$\sin A + \sin B = \frac{1}{4},$$ we have to prove that $$\cos \frac{A - B}{2} = \pm\frac{5}{24}$$

Squaring and adding $$(\cos^2A + \sin^2A) + (\cos^2B + \sin^B) + 2(\cos A\cos B + \sin A\sin B) = \frac{1}{9} + \frac{1}{16}$$

$$2 + 2\cos(A - B) = \frac{25}{144}$$

$$4\cos^2\frac{A - B}{2} = \frac{25}{144} \Rightarrow \cos\frac{A - B}{2} = \pm\frac{5}{24}$$

44. Given $$2\tan \frac{A}{2} = \tan \frac{B}{2},$$ we have to prove that $$\cos A = \frac{3 + 5\cos B}{5 + 3\cos B}$$

$$\tan\frac{A}{2} = \frac{1}{2}\tan\frac{B}{2}$$

$$\cos A = \frac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}} = \frac{1 - \frac{\tan^2\frac{B}{2}}{4}}{1 + \frac{\tan^2\frac{B}{2}}{4}}$$

$$= \frac{4 - \tan^2\frac{B}{2}}{4 + \tan^2\frac{B}{2}} = \frac{3 + 5.\frac{1 - \tan^2\frac{B}{2}}{1 + \tan^2\frac{B}{2}}}{5 + 3\frac{1 - \tan^2\frac{B}{2}}{1 + \tan^2\frac{B}{2}}}$$

$$= \frac{3 + 5\cos B}{5 + 3\cos B} =$$ R.H.S.

45. Given $$\sin A = \frac{4}{5}$$ and $$\cos B = \frac{5}{13},$$ we have to prove that one value of $$\cos \frac{A - B}{2} = \frac{8}{\sqrt{65}}$$

$$\cos A = \frac{3}{5}$$ and $$\sin B = \frac{12}{13}$$

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

$$\cos(A - B) = \cos A\cos B + \sin A\sin B = \frac{15}{65} + \frac{48}{65} = \frac{63}{65}$$

$$\frac{1 + \cos(A - B)}{2} = \frac{128}{2.65}$$

$$\cos\frac{A - B}{2} = \pm\frac{8}{\sqrt{65}}$$

46. Given, $$\sec(A + B) + \sec(A - B) = 2\sec A,$$ we have to prove that $$\cos B = \pm \sqrt{2}\cos \frac{B}{2}$$

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

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

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

$$2\cos^2A\cos B = 2\cos^2A - 1 + 2\cos^2B - 1$$

$$2\cos^2A(\cos B - 1) = 2(\cos^2B - 1)$$

$$\cos^2A = \cos B + 1 = 2\cos^2\frac{B}{2}$$

$$\cos A = \pm\sqrt{2}\cos\frac{B}{2}$$

47. Given $$\cos \theta = \frac{\cos\alpha\cos\beta}{1 - \sin\alpha\sin\beta},$$ we have to prove that one of the values of $$\tan \frac{\theta}{2}$$ is $$\frac{\tan \frac{\alpha}{2} - \tan\frac{\beta}{2}}{1 - \tan\frac{\alpha}{2}\tan\frac{\beta}{2}}$$

$$\tan^2\frac{\theta}{2} = \frac{1 - \cos\theta}{1 + \cos\theta}$$

$$= \frac{1 - \frac{\cos\alpha\cos\beta}{1 - \sin\alpha\sin\beta}}{1 + \frac{\cos\alpha\cos\beta}{1 - \sin\alpha\sin\beta}}$$

$$= \frac{1 - (\cos\alpha\cos\beta + \sin\alpha\sin\beta)}{1 + (\cos\alpha\cos\beta - \sin\alpha\sin\beta)}$$

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

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

$$\tan\frac{\theta}{2} = \frac{\sin\frac{\alpha - \beta}{2}}{\cos\frac{\alpha + \beta}{2}}$$

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

Dividing both numerator and denominator by $$\cos\frac{\alpha}{2}\cos\frac{\beta}{2}$$

$$\tan\frac{\theta}{2} = \frac{\tan \frac{\alpha}{2} - \tan\frac{\beta}{2}}{1 - \tan\frac{\alpha}{2}\tan\frac{\beta}{2}}$$

48. Given $$\tan\alpha = \frac{\sin\theta\sin\phi}{\cos\theta + \cos\phi},$$ we have to prove that one of the values of $$\tan\frac{\alpha}{2}$$ is $$\tan\frac{\theta}{2}\tan\frac{\phi}{2}$$

$$\Rightarrow \frac{2\tan\frac{\alpha}{2}}{1 - \tan^2\frac{\alpha}{2}} = \frac{\frac{2\tan\frac{\theta}{2}}{1 + \tan^2\frac{\theta}{2}}\frac{2\tan\frac{\phi}{2}}{1 + \tan^2\frac{\phi}{2}}}{\frac{1 - \tan^2\frac{\theta}{2}}{1 + \tan^2\frac{\theta}{2}} + \frac{1 - \tan^2\frac{\phi}{2}}{1 + \tan^2\frac{\phi}{2}}}$$

$$= \frac{4\tan\frac{\theta}{2}\tan\frac{\phi}{2}}{1 + \tan^2\frac{\phi}{2} - \tan^2\frac{\theta}{2} - \tan^2\frac{\phi}{2}.\tan^2\frac{\theta}{2} + 1 + \tan^2\frac{\theta}{2} -\tan^2\frac{\phi}{2} - \tan^2\frac{\phi}{2}.\tan^2\frac{\theta}{2}}$$

$$= \frac{2\tan\frac{\theta}{2}\tan\frac{\phi}{2}}{1 - \tan^2\frac{\theta}{2}\tan^2\frac{\phi}{2}}$$

Solving this quadratic equationin $$\tan\frac{\alpha}{2}$$ we obtain the desired result.

49. Given $$\cos\theta = \frac{\cos\alpha + \cos\beta}{1 + \cos\alpha\cos\beta},$$ we have to prove that one of the values of $$\tan\frac{\theta}{2}$$ is $$\tan\frac{\alpha}{2}\tan\frac{\beta}{2}$$

$$\cos\theta = \frac{\cos\alpha + \cos\beta}{1 + \cos\alpha\cos\beta}$$

$$\frac{1 - \tan^2\frac{\theta}{2}}{1 + \tan^2\frac{\theta}{2}} = \frac{\cos\alpha + \cos\beta}{1 + \cos\alpha\cos\beta}$$

$$\tan^2\frac{\theta}{2} = \frac{1 - \cos\alpha\cos\beta - \cos\alpha + \cos\beta}{1 - \cos\alpha\cos\beta + \cos\alpha - \cos\beta}$$

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

$$\tan^2\frac{\theta}{2} = \tan^2\frac{\alpha}{2}\cot^2\frac{\beta}{2}$$

$$\tan\frac{\theta}{2} = \pm\tan\frac{\alpha}{2}\cot\frac{\beta}{2}$$