5. Trigonometrical Ratios SolutionsΒΆ

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. \(= \frac{\tan A}{\sec A - 1} + \frac{\tan A}{\sec A + 1}\)

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

    \(= \frac{2\tan A\sec A}{\tan^2 A} = 2\cosec A =\) 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, \(\tan^2A + \sec A = 5 \Rightarrow \sec^2A - 1 + \sec A = 5\)

    \(\sec^2A + \secA - 6 = 0\)

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

  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 \(5\sin A - 3\cos A = x\)

    Squaring and adding, we get

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

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

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

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

    \(\Rightarrow \frac{\cos^4 A}{\cos^2 B} - \cos^2A = \sin^2A - \frac{\sin^4A}{\sin^2B} = 0\)

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

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

    \(\Rightarrow (\cos^2A - \cos^2B )\left(\frac{\cos^2A}{\cos^2B} - \frac{\sin^2A}{\sin^2B}\right) = 0\)

    When \(\cos^2A - \cos^2B = 0, \cos^2A = \cos^2B\)

    When \(\frac{\cos^2A}{\cos^2B} - \frac{\sin^2A}{\sin^2B} = 0\)

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

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

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

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

      \(= \frac{\cos^4 B}{\cos^2 B} + \frac{\sin^4 B}{\sin^2 B} = \cos^2B + \sin^2B = 1 =\) R.H.S.

  61. Given, \(\cos A + \sin A = \sqrt{2}\cos A\)

    Squaring both sides

    \(1 + 2\sin A\cos A = 2\cos^2A\)

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

    \(\cos A - \sin A = \pm \sqrt{2}\sin A\)

  62. Given, \(a\cos A - b\sin A = c,\) and let \(a\sin A + b\cos A = x\)

    Squaring and adding, we get

    \(\Rightarrow a^2\cos^2A + b^2\sin^2A - 2ab\cos A\sin A + a^2\sin^2A + b^2\cos^2A + 2ab\sin A\cos A = c^2 + x^2\)

    \(\Rightarrow a^2(\cos^2A + \sin^2A) + b^2(\sin^2A + \cos^2A) = c^2 + x^2\)

    \(\Rightarrow a^2 + b^2 = c^2 + x^2 \Rightarrow x = \pm \sqrt{a^2 + b^2 - c^2}\)

  63. Given, \(1 - \sin A = 1 + \sin A\)

    Multiplying both sides by \(1 - \sin A,\) we get

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

    \(1 - \sin A = \pm \cos A\)

    Similarly if we multiply with \(1 + \sin\) we can prove that

    \(1 + \sin A = \pm \cos A\)

  64. Let us solve these one by one:

    1. Given, \(\sin^4A + \sin^2A = 1 \Rightarrow \sin^4A = 1 - \sin^2A = \cos^2\)

      \(\frac{\sin^4A}{\cos^4A} = sec^2A\)

      \(\frac{1}{\tan^4A} = cos^2A\)

      Also from, \(\sin^4A = \cos^2A \Rightarrow \tan^2A = \cosec^2A \Rightarrow \frac{1}{\tan^2A} = \sin^2A\)

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

    2. Given, \(\sin^4A + \sin^2A = 1 \Rightarrow \sin^4A = 1 - \sin^2A = \cos^2A\)

    \(\Rightarrow \frac{\sin^2A}{\cos^2A} = \frac{1}{\sin^2A} \Rightarrow \tan^2A = \cosec^2A\)

    \(\tan^2A = 1 + \cot^2A\)

    Multiplying both sides with \(\tan^2A,\) we get

    \(\tan^4A = \tan^2A + 1\)

    \(\tan^4A - \tan^2A = 1\)

  65. Given, \(\cos^2 - \sin^2 A = \tan^2 B \Rightarrow \frac{\cos^2A - \sin^2A}{\cos^2A + \sin^2A} = \tan^2B\)

    Dividing both numerator and denominator of L.H.S. with \(\cos^2A,\) we get

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

    \(1 - \tan^2A - \tan^2B - \tan^2A\tan^2B = 0\)

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

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

  66. We will prove this by induction. Let \(\sin A + \cosec A = 2,\) thus it is true for \(n = 1\)

    Squaring both sides \(\sin^2A + \cosec^2A + 2\sin A\cosec A = 2^2 \Rightarrow \sin^2A + \cosec^2A = 2\)

    Thus, it is true for \(n = 2\) as well. Let it be true for \(n = m-1\) and \(n = m\)

    \(\sin^mA + \cosec^mA = 2\)

    Multiplying both sides with \(\sin A + \cosec A,\) we get

    \(\sin^{m + 1}A + \cosec^{m + 1}A + \sin^mA\cosec A + \cosec^mA\sin A = 2^2\)

    \(\sin^{m + 1}A + \cosec^{m + 1}A + \sin^{m - 1}A + \cosec^{m - 1}A = 4\)

    \(\sin^{m + 1}A + \cosec^{m + 1} = 4 - 2 = 2\)

    Thus, we have proven it by induction.

  67. L.H.S. \(= \sec A + \tan^3A\cosec A = \sec A\left(1 + \tan^3A\frac{\cosec A}{\sec A}\right)\)

    \(=\sec A(1 + \tan^3A\cot a) = \sec A(1 + \tan^2A) = \sec^3A = (1 + 1 - e^2)\frac{3}{2} = (2 - e^2)\frac{3}{2}\)

  68. Cross multiplying given equations, we have

    \(\frac{\sec A}{br - qc} = \frac{\tan A}{pc - qr} = \frac{1}{qa - pb}\)

    \(\therefore \sec^2A - \tan^2A = 1, \left(\frac{br - qc}{aq - pb}\right)^2 - \left(\frac{pc - qa}{aq - pb}\right)^2 = 1\)

    \((br - qa)^2 - (pc - ar)^2 = (qa - pb)^2\)

  69. Given, \(\cosec A - \sin A = m \Rightarrow \frac{1}{\sin A } - \sin A = m\)

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

    Also given, \(\sec A - \cos A = n \Rightarrow \frac{1}{\cos A} - \cos A = n\)

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

    We have \(\sin A = \frac{\cos^2A}{m},\) putting this in derived equation

    \(\cos^3A = m^2n\)

    \(\therefore \sin A = \frac{(m^2n)^\frac{2}{3}}{m} = (mn^2)\frac{1}{3}\)

    Thus, \(\sin^2A + \cos^2A = 1\) gives us \((m^2n)^\frac{2}{3} + (mn^2)^\frac{2}{3} = 1\)

  70. Given, \(\sec^2A = \frac{4xy}{(x + y)^2}\)

    \(\because \sec^2A \geq 1 \therefore \frac{4xy}{(x + y)^2}\geq 1\)

    \(\Rightarrow (x + y)^2 \leq 4xy \Rightarrow (x - y)^2 \leq 0\)

    But for real \(\x\) and \(y, (x - y)^2\nless 0\)

    \(\therefore (x - y)^2 = 0 \therefore x - y\)

    Also, \(x + y \neq 0 \Rightarrow x\neq 0, y \neq 0\)

  71. Given, \(\sin A = x + \frac{1}{x},\) squaring we get

    \(\sin^2A = x^2 + \frac{1}{x^2} + 2 \geq 2\)

    which is not possible since \(\sin A \leq 1\)

  72. Given, \(\sec A - \tan A = p\)

    \(\Rightarrow 1 - \sin A = p\cos A\)

    Squaring, we obtain

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

    \((1 + p^)\sin^2A - 2\sin A + 1 - p^2 = 0\)

    \(\sin A = \frac{1 \pm p^2}{1 + p^2}\)

    Now \(tan\) and \(\sec A\) can be easily found.

  73. \(\sec A + \tan A = \frac{4p^2 + 1}{4p} \pm \sqrt{\sec^2A - 1}\)

    \(= 4p \pm \sqrt{\left(\frac{(4p^2 + 1)^2}{16p^2} - 1\right)}\)

    \(= 2p\) or \(\frac{1}{2p}\)

  74. Given, \(\frac{\sin A}{\sin B} = p, \frac{\cos A}{\cos B} = q\)

    Squaring, we get

    \(\frac{\sin^2A}{\sin^2B} = p^2, \frac{\cos^2A}{\cos^2B}= q^\)

    \(\frac{\sin^2A - \sin^2B}{\sin^2B} = p^2 - 1, \frac{\cos^2B - \cos^2A}{\cos^2B} = 1 - q^2\)

    \(\frac{\sin^2A - \sin^2B}{\sin^2B} = p^2 - 1, \frac{\sin^2A - \sin^2B}{\cos^2B} = 1 - q^2\)

    Dividing, we obtain

    \(\tan^2B = \pm \sqrt{\frac{1 - q^2}{p^2 - 1}}\)

    Dividing original equations

    \(\frac{\tan A}{\tan B} = \pm \frac{p}{q}\sqrt{\frac{1 - q^2}{p^2 - 1}}\)

  75. \(\sin A = \sqrt{2}\sin B, \frac{\sin A.\cos B}{\cos A. \sin B} = \sqrt{3}\)

    Substituting for \(\sin A\)

    \(\frac{\sqrt{2}\sin B\cos B}{\sqrt{1 - 2\sin^2A}\sin B} = \sqrt{3}\)

    Squaring, we get

    \(2\cos^2 A = 3(1 - 2\sin^2A) \Rightarrow 4\sin^2A - 1 = 0\)

    \(\Rightarrow \sin A = \pm \frac{1}{2}, \Rightarrow A = \pm 45^\circ\)

    Thus, \(B = \pm 30^\circ\)

  76. Given, \(\tan A + \cot A = 2, 1 + \tan^2A = 2\tan A \Rightarrow (1 - \tan A)^2 = 0\)

    \(\tan A = 1\Rightarrow \sin A = \pm\frac{1}{\sqrt{2}}\)

  77. \(m^2 - n^2 = (\tan A +\sin A)^2 - (\tan A - \sin A)^2 = 4\tan A\sin A\)

    \(4\sqrt{mn} = 4\sqrt{\tan^2A - \sin^2A} = 4\sqrt{\frac{\sin^2A}{\cos^2A} - \sin^2A}\)

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

    \(= 4\tan A\sin A\)

  78. \(m^2 - 1 = 2\sin A\cos A \Rightarrow n(m^2 - 1) = 2\sin A\cos A(\sec A + \cosec A)\)

    \(= 2\sin A + 2\cos A = 2m\)

  79. Given, \(x\sin^3 A + y\cos^3 A = \sin A\cos A\)

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

    Frpm other equation \(x\sin A = y\cos A,\) substituting this in above equation

    \((x\sin A)\sin^2A + (x\sin A)\cos^2a = \sin A\cos A\)

    \(x\sin A = \sin A\cos A\)

    \(x = \cos A \therefore y = \sin A\)

    Thus, \(x^2 + y^2 = 1\)

  80. Given, \(\sin^2A = \frac{(x + y)^2}{4xy}\)

    \(\sin^2A \leq 1 \Rightarrow (x + y)^2 \leq 4xy \Rightarrow (x - y)^2 \leq 0\)

    But for real \(x\) and \(y\) \((x - y)^2\nless 0\)

    \(\Rightarrow x = y\)

    Also, \(xy \neq 0, x, y\neq 0\)