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