# 5. Trigonometrical Ratios Solutions¶

Go through the list of relationship between trigonometrical ratios as the problems are centered around those.

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

Multiplying and dividing with $1 - \cos A,$ we get

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

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

$= \frac{1 - \cos A}{\sin A} = \cosec A - \cot A =$ R.H.S.

2. L.H.S. $\sqrt{1 + \tan^2A + 1 + \cot^2A}$ [$\because \sec^2A = 1 + \tan^2A$ and $\cosec^2A = 1 + \cot^2 A$ ]

$= \sqrt{\tan^2A + 2\tan A\cot A + \cot^2A}$ [$\tan A\cot A = 1$ ]

$= \sqrt{(\tan A + \cot A)^2} =$ R.H.S.

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

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

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

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

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

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

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

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

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

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

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

8. L.H.S. $= \sqrt{\frac{1 - \sin A}{1 + \sin A}}$

Multiplying and dividing with $1 - \sin A,$ we get

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

$= \frac{1 - \sin A}{\cos A} = \sec A - \tan A$

9. L.H.S. $= \frac{\cosec A}{\cosec A - 1} + \frac{\cosec A}{\cosec A + 1} = \frac{\frac{1}{\sin A}}{\frac{1}{\sin A} - 1} + \frac{\frac{1}{\sin A}}{\frac{1}{\sin A} + 1}$

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

10. L.H.S. $= \frac{\cosec A}{\tan A + \cot A} = \frac{\frac{1}{\sin A}}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$

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

11. L.H.S. $= (\sec A + \cos A)(\sec A - \cos A) = \sec^2A - \cos^2A = 1 + \tan^2A - 1 + \sin^2A = \tan^2A + \sin^2A =$ R.H.S.

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

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

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

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

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

Multiplying and dividing with $\sec A - \tan A,$ we get

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

$= 1 - 2\tan A\sec A + 2\tan^2A =$ R.H.S.

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

Multiplying and dividing with $\sec A + \tan A,$ we get

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

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

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

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

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

$= \frac{1 + \sin A \cos A}{\sin A\cos A} = 1 + \cosec A\sec A =$ R.H.S.

18. L.H.S. $= \frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \frac{\cos A}{1 - \frac{\sin A}{\cos A}} + \frac{\sin A}{1 - \frac{\cos A}{\sin A}}$

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

$= \cos A + \sin A =$ R.H.S.

19. L.H.S. $= (\sin A + \cos A)(\tan A + \cot A) = (\sin A + \cos A)(\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A})$

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

20. L.H.S. $= \sec^4A - \sec^2A = (1 + \tan^2A) - 1 - \tan^2A = 1 + 2\tan^2A + \tan^4A - 1 - \tan^2A = \tan^4A + \tan^2A =$ R.H.S.

21. L.H.S. $= \cot^4A + \cot^2A = (\cosec^2A - 1)^1 + \cosec^2A - 1 = \cosec^4A - 2\cosec^2A + 1 + \cosec^2A -1 = \cosec^4A - \cosec^2A =$ R.H.S.

22. L.H.S. $= \sqrt{\cosec^2A - 1} = \sqrt{\cot^2A} = \cot A = \frac{\cos A}{\sin A} = \cos A \cosec A =$ R.H.S.

23. L.H.S. $= \sec^2A\cosec^2A = (1 + \tan^2A)(1 + \cot^2A) = 1 + \tan^2A + \cot^2A + \tan^2A\cos^2A = 2 + \tan^2A + \cot^2A =$ R.H.S.

24. L.H.S. $= \tan^2A - \sin^2A = \frac{\sin^2A}{\cos^2A} - \sin^2A = \frac{\sin^2A - \sin^2A\cos^2A}{\cos^2A} = \sin^2A(1 - \cos^2A)\sec^2A = \sin^4A\sec^2A =$ R.H.S.

25. L.H.S. $= (1 + \cot A - \cosec A)(1 + \tan A + \sec A) = \left(1 + \frac{\cos A}{\sin A} - \frac{1}{\sin A}\right)\left(1 + \frac{\sin A}{\cos A} + \frac{1}{\cos A}\right)$

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

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

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

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

Proceeding in same way for R.H.S. we obtain $\frac{1 - \sin A}{\cos A}$

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

27. L.H.S. $= \frac{\cot A + \tan B}{\cot B + \tan A} = \frac{\frac{1}{\tan A} + \tan B}{\frac{1}{\tan B} + \tan A}$

$\frac{\frac{1 + \tan A\tan B}{\tan A}}{\frac{1 + \tan A\tan B}{\tan B}} = \frac{\tan B}{\tan A} = \cot A\tan B =$ R.H.S.

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

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

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

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

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

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

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

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

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

29. L.H.S. $= \sin^8A - \cos^8A = (\sin^4A - \cos^4A)(\sin^4A + \cos^4A)$

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

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

30. L.H.S. $= \frac{\cos A\cosec A - \sin A\sec A}{\cos A + \sin A} = \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\cos A}}{\cos A + \sin A}$

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

31. Given, $\frac{1}{\cosec A - \cot A} - \frac{1}{\sin A} = \frac{1}{\sin A} - \frac{1}{\cosec A + \cot A}$

Therefore alternatively we can prove that $\frac{1}{\cosec A - \cot A} + \frac{1}{\cosec A + \cot A} = \frac{2}{\sin A}$

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

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

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

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

Solving this yields $= \frac{1 + \sin A}{\cos A} =$ R.H.S.

33. L.H.S. $= (\tan A + \cosec B)^2 - (\cot B - \sec A)^2 = \tan^2A + \cosec^2B + 2\tan A\cosec B - \cot^2B - \sec^2A+ 2\sec A\cot B$

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

$= 1 - 1 + 2\tan A\cot B\left(\sec B + \cosec A\right) =$ R.H.S.

34. L.H.S. $= 2\sec^2 A - \sec^4A - 2\cosec^2A + \cosec^4A = 2(1 + \tan^2A) - (1 + \tan^2A)^2 - 2(1 + \cot^2A) + (1 + \cot^2A)^2$

$= 2 + 2\tan^2A - 1 - 2\tan^2A + \tan^4A - 2 - 2\cot^2A + 1 + 2\cot^2A + \cot^4A = \cot^4A - \tan^4A =$ R.H.S.

35. L.H.S. $= (\sin A + \cosec A)^2 + (\cos A + \sec A)^2 = \sin^2A + \cosec^2A + 2\sin A\cosec A + \cos^2A + \sec^2A + 2\sec A\cosec A$

$= (\sin^2A + \cos^2A) + 4 + \cosec^2A + \sec^2A = 1 + 4 + 1 + \cot^2A + 1 + \tan^2A = \cot^2A + \tan^2A + 7 =$ R.H.S.

36. L.H.S $= (\cosec A + \cot A)(1 - \sin A) - (\sec A + \tan A)(1 - \cos A)$

$= \left(\frac{1}{\sin A} + \frac{\cos A}{\sin A}\right)(1 - \sin A) - \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)(1 - \cos A)$

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

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

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

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

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

$= (\cosec A - \sec A)(2 - (1 - \cos A)(1 - \sin A)) =$ R.H.S.

37. L.H.S $= (1 + \cot A + \tan A)(\sin A - \cos A) = \left(1 + \frac{\cosec A}{\sec A} + \frac{\sec A}{\cosec A}\right)\left(\frac{1}{\cosec A} - \frac{1}{\sec A}\right)$

$= \left(\frac{\sec A\cosec A + \sec A | \cosec A}{\sec A\cosec A}\right)\left(\frac{\sec A - \cosec A}{\sec A\cosec a}\right)$

$= \frac{\sec^3A - \cosec^3A}{\sec^2A\cosec^2A} = \frac{\sec A}{\cosec^2A} - \frac{\cosec A}{\sec^2A} =$ R.H.S.

38. Given, $\frac{1}{\sec A - \tan A} - \frac{1}{\cos A} = \frac{1}{\cos A} - \frac{1}{\sec A + \tan A}$

So alternatively we can prove that $\frac{1}{\sec A - \tan A} + \frac{1}{\sec A + \tan A} = \frac{2}{\cos A}$

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

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

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

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

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

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

$= 3 + 12\sin^2A\cos^2A -12\cos A\sin A + 4(1 - 3\sin^2A\cos^2A) + 6 + 12\sin A\cos A$

$= 13 =$ R.H.S.

40. L.H.S. $= \sqrt{\frac{1 + \cos A}{1 - \cos A}}$

Multiplying and dividing with $1 + \cos A,$ we get

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Hence, L.H.S. = R.H.S.

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

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

$= \tan^4A + sec^4A =$ R.H.S.

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

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

1. $= \frac{\cot A + \cosec A - 1}{\cot A - \cosec A + 1}$

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

Multiplying and dividing with $\cos A - (1 - \sin A),$ we get

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

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

Multiplyig numerator and denominator with $1 + \cos A,$ we get

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

47. L.H.S. $= (\sin A + \sec A)^2 + (\cos A + \cosec A)^2$

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

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

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

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

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

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

$= \frac{2\sin A}{1 + \tan A} =$ R.H.S.

49. Given, $2\sin A = 2 - \cos A$

$\cos A = 2 - 2\sin A,$ squaring both sides we, get $\cos^2A = 4 - 8\sin A + 4\sin^2A$

$1 - \sin^2A = 4 - 8\sin A + 4\sin^2A \Rightarrow 5\sin^2A - 8\sin A + 3 = 0$

$5\sin^2A - 5\sin A - 3\sin A + 3 = 0 \Rightarrow \sin A = 1, \frac{3}{5}$

50. Given $8\sin A = 4 + \cos A \Rightarrow 8\sin A - 4 = \cos A$

Squaring both sides, we get

$64\sin^2A - 64\sin A + 16 = \cos^2A = 1 - \sin^2A$

$65\sin^2A - 64\sin A + 15 = 0$

$65\sin^2A - 39\sin A - 25\sin A + 15 = 0$

$\sin A = \frac{3}{5}, \frac{5}{13}$

51. Given, $\tan A + \sec A = 1.5$

$\Rightarrow 1 + \sin A = 1.5 \cos A \Rightarrow 2 + 2\sin A = 3\cos A$

Squaring both sides, we get

$4 + 8\sin A + 4\sin^2A = 9 - 9\sin^2A$

$13\sin^2A + 8\sin A - 5 =0$

$\sin A = -1, \frac{5}{13}$

52. Given, $\cot A + \cosec A = 5 \Rightarrow 1 + \cos A = 5\sin A$

Squarig both sides, we get

$1 + 2\cos A + \cos^2A = 25\sin^2A = 25(1 - \cos^2A)$

$26\cos^2A + 2\cos A - 24 = 0$

$26\cos^2A + 26\cos A - 24\cos A - 24 = 0$

$\cos A= -1, \frac{12}{13}$

53. $3\sec^4 A + 8 = 10\sec^2A \Rightarrow 3(sec^2A)^2 + 8 = 10(1 + \tan^2A)$

$\Rightarrow 3(1 + \tan^2A)^2 + 8 = 10 + 10\tan^2A$

$3 + 6\tan^2A + 3\tan^4A + 8 = 10 + 10\tan^2A$

$3\tan^4A - 4\tan^2A + 1 = 0$

$\tan A = \pm1, \pm\frac{1}{\sqrt{3}}$

54. Given, $\tan^2A + \sec A = 5 \Rightarrow \sec^2A - 1 + \sec A = 5$

$\sec^2A + \sec A - 6 = 0$

$\sec A = -3, 2\Rightarrow \cos A = -\frac{1}{3}, \frac{1}{2}$

55. Given $\tan A + \cot A = 2 \Rightarrow \tan A + \frac{1}{\tan A} = 2$

$\tan^2A - 2\tan A + 1 = 0$

$\tan A = 1 \Rightarrow \sin A = \frac{1}{\sqrt{2}}$

56. Given, $\sec^2A = 2 + 2\tan A \Rightarrow \tan^2A + 1 = 2 + 2\tan A$

$\tan^2A - 2\tan A - 1 = 0$

$\tan A = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$

57. Given, $\tan A = \frac{2x(x + 1)}{2x + 1}$

$\sin A = \frac{2x(x + 1)}{\sqrt{[2x(x + 1)]^2 + (2x + 1)^2}}$

$\cos A = \frac{2x + 1}{\sqrt{[2x(x + 1)]^2 + (2x + 1)^2}}$

58. Given, $3\sin A + 5\cos A = 5,$ let $5\sin A - 3\cos A = x$

$9\sin^2A + 25\cos^2A + 30\sin A\cos A + 25\sin^2A + 9\cos^2A - 30\sin A\cos A = 25 + x^2$

$9(\sin^2A + \cos^2A) + 25(\cos^2A + \sin^2A) = 25 + x^2$

$34 = 25 + x^2 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$

59. Given, $\sec A + \tan A = \sec A - \tan A$

Multiplying both sides with $\sec A + \tan A,$ we get

$(\sec A + \tan A)^2 = \sec^2A - \tan^2A = 1$

$\sec A + \tan A = \pm 1$

We can prove that $\sec A - \tan A$ to be $\pm 1$ by multiplying given equation with $\sec A - \tan A$

60. Given, $\frac{\cos^4 A}{\cos^2 B} + \frac{\sin^4 A}{\sin^2 B} = 1 = \sin^2A + \cos^2A$