9. Compound Angles SolutionsΒΆ

  1. Given, \(\sin\alpha = \frac{3}{5}\) and \(\cos\beta = \frac{9}{41}\)

    Therefore, \(\cos\alpha = \frac{4}{5}\) and \(\sin\beta = \frac{40}{41}\)

    \(\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta\)

    \(= \frac{3}{5}\frac{9}{41} - \frac{4}{5}\frac{40}{41}\)

    \(= \frac{27 - 160}{205} = -\frac{133}{205}\)

    \(\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta\)

    \(= \frac{4}{5}\frac{9}{41} - \frac{3}{5}\frac{40}{41}\)

    \(= \frac{36 - 120}{205} = -\frac{84}{205}\)

  2. Given, \(\sin\alpha = \frac{45}{53}\) and \(\sin\beta = \frac{33}{65}\)

    Thus, \(\cos\alpha = \frac{28}{53}\) and \(\cos\beta = \frac{56}{65}\)

    \(\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta\)

    \(= \frac{45}{53}\frac{56}{65} - \frac{28}{53}\frac{33}{65}\)

    \(= \frac{2520 - 924}{3445} = \frac{1596}{3445}\)

    \(\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta\)

    \(= \frac{45}{53}\frac{56}{65} + \frac{28}{53}\frac{33}{65}\)

    \(= \frac{2520 + 924}{3445} = \frac{3444}{3445}\)

  3. Given, \(\sin\alpha = \frac{15}{17}\) and \(\cos\beta = \frac{12}{13}\)

    \(\cos\alpha = \frac{8}{17}\) and \(\sin\beta = \frac{5}{13}\)

    \(\tan\alpha = \frac{15}{8}\) and \(\tan\beta = \frac{5}{12}\)

    \(\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta\)

    \(= \frac{15}{17}\frac{12}{13} + \frac{8}{17}\frac{5}{13}\)

    \(= \frac{220}{221}\)

    \(\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta\)

    \(= \frac{8}{17}\frac{12}{13} + \frac{15}{17}\frac{5}{13}\)

    \(= \frac{171}{221}\)

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

    \(= \frac{\frac{15}{8} + \frac{5}{12}}{1 - \frac{15}{8}\frac{5}{12}}\)

    \(= \frac{220}{21}\)

  4. L.H.S. \(= \cos(45^{\circ} - A)\cos(45^{\circ} - B) - \sin(45^{\circ} - A)\sin(45^{\circ} - B)\)

    \(= [(\cos 45^\circ\cos A + \sin45^\circ\sin A)(\cos 45^\circ\cos B + \sin45^\circ\sin B) - (\sin45^\circ\cos A - \cos45^\circ\sin A)(\sin45^\circ\cos B - \cos45^\circ\sin B)]\)

    Substituting valus for \(\sin45^\circ\) and \(\cos45^\circ\)

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

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

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

  5. L.H.S. \(= \sin(45^{\circ} + A)\cos(45^\circ - B) + \cos(45^{\circ} + A)\sin(45^\circ - B)\)

    \(= [(\sin45^\circ\cos A + \cos45^\circ\sin A)(\cos45^\circ\cos B + \sin45^\circ\sin B) + (\cos45^\circ\cos A - \sin45^\circ\sin A)(\sin45^\circ\cos B - \cos45^\circ)\sin B]\)

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

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

    \(= \cos A\cos B + \sin A\sin B = \cos(A - B)\)

  6. L.H.S. \(= \frac{\sin(A - B)}{\cos A\cos B} + \frac{\sin(B - C)}{\cos B\cos C} + \frac{\sin(C - A)}{\cos C\cos A}\)

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

    \(= \tan A - \tan B + \tan B - \tan C + \tan C - \tan A = 0 =\) R.H.S.

  7. L.H.S. \(= \sin 105^\circ + \cos 105^\circ\)

    \(= \sin(60^\circ + 45^\circ) + \cos(60^\circ + 45^\circ)\)

    \(=\sin60^\circ\cos45^\circ + \cos60^\circ\sin45^\circ + \cos60^\circ\cos45^\circ - \sin60^\circ\sin45^\circ\)

    \(=\cos45^\circ(\sin60^\circ + \cos60^\circ + \cos60^\circ - \sin 60^\circ)[\because \sin45^\circ = \cos45^\circ]\)

    \(= \cos45^\circ =\) R.H.S.

  8. Given, \(\sin 75^\circ - \sin 15^\circ = \cos 105^\circ + \cos 15^\circ\)

    \(\Rightarrow \sin75^\circ -\sin15^\circ = \cos(90^\circ + 15^\circ) + \sin(90^\circ - 15^\circ)\)

    \(\Rightarrow \sin75^\circ -\sin15^\circ = \cos90^\circ\cos15^\circ - \sin90^\circ\sin15^\circ + \sin(90^\circ - 15^\circ)\)

    \(\Rightarrow \sin75^\circ -\sin15^\circ = -\sin15^\circ + \sin75^\circ [\because \cos90^\circ = 0~\&~\sin90^\circ = 1]\)

    Thus, we have proven the equality.

  9. L.H.S. \(= \cos\alpha\cos(\gamma - \alpha) - \sin\alpha\sin(\gamma - \alpha)\)

    \(= \cos\alpha(\cos\gamma\cos\alpha + \sin\gamma\sin\alpha) - \sin\alpha(\sin\gamma\cos\alpha - \sin\alpha\cos\gamma)\)

    \(= \cos^2\alpha\cos\gamma + \sin\gamma\sin\alpha\cos\alpha - \sin\alpha\sin\gamma\cos\alpha - \sin^2\alpha\cos\gamma\)

    \(= \cos\gamma(\sin^2\alpha + \cos^2\alpha) = \cos\gamma =\) R.H.S.

  10. L.H.S. \(= \cos(\alpha + \beta)\cos\gamma - \cos(\beta + \gamma)\cos\alpha\)

    \(= \cos\alpha\cos\beta\cos\gamma - \sin\alpha\sin\beta\cos\gamma - \cos\alpha\cos\beta\cos\gamma + \cos\alpha\sin\beta\sin\gamma\)

    \(= \sin\beta(\cos\alpha\sin\gamma - \sin\alpha\cos\gamma)\)

    \(= \sin\beta\sin(\gamma - \alpha) =\) R.H.S.

  11. L.H.S. \(= \sin(n + 1)A\sin(n - 1)A + \cos(n + 1)A\cos(n - 1)A\)

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

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

    \(= \cos(n + 1 - (n + 1)) = \cos A =\) R.H.S.

  13. \(\cos 15^\circ = \cos(45^\circ - \cos 30^\circ)\)

    \(= \cos45^\circ\cos30^\circ + \sin45^\circ\sin30^\circ\)

    \(= \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\frac{1}{2}\)

    \(= \frac{\sqrt{3} + 1}{2\sqrt{2}}\)

    \(\sin 105^\circ = \sin(60^\circ + 45^\circ)\)

    \(= \sin60^\circ\cos^45\circ + \cos60^\circ\sin45^\circ\)

    \(= \frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}} + \frac{1}{2}\frac{1}{\sqrt{2}}\)

    \(= \frac{\sqrt{3} + 1}{2\sqrt{2}} =\) R.H.S.

  14. \(\tan 105^\circ = \tan(60^\circ + 45^\circ)\)

    \(= \frac{\tan60^\circ + \tan 45^\circ}{1 - \tan60^\circ\tan 45^\circ}\)

    \(= \frac{\sqrt{3} + 1}{1 - \sqrt{3}}\)

  15. \(\frac{\tan 495^\circ}{\cot 855^\circ} = \frac{\tan(360^\circ + 135^\circ)}{\cot(720^\circ + 135^\circ)}\)

    \(= \frac{\tan 135^\circ}{\cot 135^\circ} = \tan^2135^\circ = (-1)^2 = 1\)

  16. \(\sin(\pi + \theta) = -\sin\theta \therefore \sin(n\pi + \theta) = (1)^n\sin\theta\)

    \(\sin\left(n\pi + (-1)^n \frac{\pi}{4}\right) = (-1)^n\sin\left((-1)^n\frac{\pi}{4}\right)\)

    \(= (-1)^n(-1)^n\sin \frac{\pi}{4}~[\because \sin(-\theta) = -\sin\theta]\)

    \(= (-1)^{2n}\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}\)

  17. L.H.S. \(= \sin 15^\circ = \sin(60^\circ - 45^\circ)\)

    \(= \sin60^\circ\cos45^\circ - \cos60^\circ\sin45^\circ\)

    \(= \frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}} - \frac{1}{2}\frac{1}{\sqrt{2}}\)

    \(= \frac{\sqrt{3} - 1}{2\sqrt{2}} =\) R.H.S.

  18. L.H.S. \(= \cos 75^\circ = \cos(45^\circ + 30^\circ)\)

    \(= \cos45^\circ\cos30^\circ - \sin45^\circ\sin30^\circ\)

    \(= \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\frac{1}{2}\)

    \(= \frac{\sqrt{3} - 1}{2\sqrt{2}} =\) R.H.S.

  19. L.H.S. \(= \tan 75^\circ = \tan(45^\circ + 30^\circ)\)

    \(= \frac{\tan45^\circ + \tan30^\circ}{1 - \tan45^\circ\tan30^\circ}\)

    \(= \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1.\frac{1}{\sqrt{3}}}\)

    \(= \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}}\)

    \(= \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = 2 + \sqrt{3} =\) R.H.S.

  20. L.H.S. \(= \tan 15^\circ = \tan(45^\circ - 30^\circ)\)

    \(= \frac{\tan45^\circ - \tan30^\circ}{1 + \tan45^\circ\tan30^\circ}\)

    \(= \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1.\frac{1}{\sqrt{3}}}\)

    \(= \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}}\)

    \(= \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = 2 - \sqrt{3} =\) R.H.S.

  21. \(\cos 1395^\circ = \cos(3*360^\circ + 315^\circ) = \cos315^\circ = \cos(270^\circ + 45^\circ)\)

    \(= \cos45^\circ = \frac{1}{\sqrt{2}}\)

  22. \(\tan(-330^\circ) = -\tan(330^\circ) = -\tan(270^\circ + 60^\circ) = =\cot 60^\circ = \frac{1}{\sqrt{3}}\)

  23. Given, \(\sin 300^\circ \cosec 1050^\circ - \tan(-120^\circ)\)

    \(= \sin (270^\circ + 30^\circ)\cosec(720^\circ + 270^\circ + 60^\circ) + \tan(90^\circ + 30^\circ)\)

    \(= -cos 30^\circ. \frac{1}{-cos 60^\circ} - \cot 30^\circ\)

    \(= \frac{\sqrt{3}}{2}.\frac{2}{1} - \sqrt{3} = 0\)

  24. Given, \(\tan\left(\frac{11\pi}{12}\right)\)

    \(= \tan\left(\pi - \frac{\pi}{12}\right) = -\tan15^\circ\)

    Using the value computed in 20 for \(\tan15^\circ\) we have \(\sqrt{3} - 2\) as the answer.

  25. We know that \(tan(-\theta) = -\tan\theta,\) thus

    \(\tan \left((-1)^n\frac{\pi}{4}\right) = (-1)^n\tan\frac{\pi}{4} = (-1)^n\)

  26. Given, \(\cos 18^\circ - \sin 18^\circ = \sqrt{2}\sin 27^\circ\)

    \(\frac{1}{\sqrt{2}}\cos\18^\circ - \frac{1}{\sqrt{2}}\sin18^\circ = \sin 27^\circ\)

    L.H.S. \(= \sin45^\circ\cos18^\circ - \cos45^\circ\sin18^\circ\)

    \(= \sin(45^\circ - 18^\circ) = \sin 27^\circ =\) R.H.S.

  27. L.H.S. \(=\tan 70^\circ = \tan(50^\circ + 20^\circ)\)

    \(= \frac{\tan 50^\circ + \tan 20^\circ}{1 - \tan50^\circ\tan20^\circ}\)

    \(\tan70^\circ - \tan70^\circ\tan50^\circ\tan20^\circ = \tan 50^\circ + \tan 20^\circ\)

    \(\tan70^\circ = \tan70^\circ\tan50^\circ\tan20^\circ + \tan 50^\circ + \tan 20^\circ\)

    \(= \tan(90^\circ - 20^\circ)\tan50^\circ\tan20^\circ + \tan 50^\circ + \tan 20^\circ\)

    \(= \cot20^\circ\tan50^\circ\tan20^\circ + \tan 50^\circ + \tan 20^\circ\)

    \(= \tan 50^\circ + \tan 50^\circ + \tan 20^\circ = 2\tan50^\circ + \tan20^\circ =\) R.H.S.

  28. L.H.S. \(= \frac{\cos\left(\frac{\pi}{4} + x\right)\cos\left(\frac{\pi}{4} - x\right)}{\sin\left(\frac{\pi}{4} + x\right)\sin\left(\frac{\pi}{4} - x\right)}\)

    \(= \frac{\cos^2\frac{\pi}{4} - \sin^2x}{\sin^2\frac{\pi}{4} - \sin^2x} = \frac{\frac{1}{2} - \sin^2x}{\frac{1}{2} - \sin^2x} = 1 =\) R.H.S.

  29. L.H.S. \(= \cos(m + n)\theta.\cos(m - n)\theta - \sin(m + n)\theta\sin(m - n)\theta\)

    \(= \cos(m + n + m - n)\theta = \cos2m\theta =\) R.H.S.

  30. L.H.S. \(= \frac{\tan(\theta + \phi) + \tan(\theta - \phi)}{1 - \tan(\theta + \phi)\tan(\theta - \phi)}\)

    \(= \tan(\theta + \phi + \theta - \phi) = \tan 2\theta =\) R.H.S.

  31. Given \(\cos 9^\circ + \sin 9^\circ = \sqrt{2}\sin 54^\circ\)

    \(\frac{1}{\sqrt{2}}\cos9^\circ + \frac{1}{\sqrt{2}}\sin9^\circ = \sin54^\circ\)

    L.H.S. \(= \sin45^\circ\cos9^\circ + \cos45^\circ\sin9^\circ\)

    \(= \sin(45^\circ + 9^\circ) = \sin 54^\circ =\) R.H.S.

  32. L.H.S. \(= \frac{\cos 20^\circ - \sin 20^\circ}{\cos 20^\circ + \sin 20^\circ}\)

    Dividing both numerator and denominaor with \(\cos20^\circ,\) we get

    \(= \frac{1 - \tan20^\circ}{1 + \tan20^\circ} = \frac{\tan 45^\circ - \tan20^\circ}{1 - \tan45^\circ\tan20^\circ}~[\because \tan45^\circ = 1]\)

    \(= \tan(45^\circ - 20^\circ) = \tan 25^\circ =\) R.H.S.

  33. L.H.S. \(= \frac{\tan A + \tan B}{\tan A - \tan B}\)

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

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

  34. L.H.S. \(= \frac{1}{\tan 3A - \tan A} - \frac{1}{\cot 3A - \cot A}\)

    \(= \frac{1}{\tan 3A - \tan A} - \frac{1}{\frac{1}{\tan 3A} - \frac{1}{\tan A}}\)

    \(= \frac{1}{\tan 3A - \tan A} - \frac{\tan A\tan 3A}{\tan A - \tan 3A}\)

    \(= \frac{1 + \tan A \tan 3A}{\tan 3A - \tan A} = \frac{1}{\tan(3A - A)} = \cot 2A =\) R.H.S.

  35. This is similar to previous problema and can be solved likewise.

  36. L.H.S. \(= \frac{\sin 3\alpha}{\sin\alpha} + \frac{\cos 3\alpha}{cos\alpha}\)

    \(= \frac{\sin3\alpha\cos\alpha + \cos3\alpha\sin\alpha}{\sin\alpha\cos\alpha}\)

    \(= \frac{\sin(3\alpha + \alpha)}{\sin\alpha\cos\alpha}\)

    \(=2\frac{2\sin 4\alpha}{\sin2\alpha}~[\because \sin\alpha\cos\alpha = \frac{1}{2}\sin2\alpha]\)

    \(= 2\frac{2\sin2\alpha\cos2\alpha}{\sin2\alpha} = 4\cos2\alpha =\) R.H.S.

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

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

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

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

  38. Given, \(\tan 40^\circ + 2 \tan 10^\circ = \tan 50^\circ\)

    R.H.S. \(= \tan 50^\circ = \tan(40^\circ + 10^\circ)\)

    \(= \frac{\tan 40^\circ + \tan 10^\circ}{1 - \tan 40^\circ\tan 10^\circ}\)

    \(\tan50^\circ - \tan50^\circ\tan40^\circ\tan10^\circ = \tan 40^\circ + \tan 10^\circ\)

    \(\tan50^\circ - \cot40^\circ\tan40^\circ\tan10^\circ = tan40^\circ + \tan10^\circ\)

    \(\tan50^\circ = \tan 40^\circ + 2 \tan 10^\circ\)

  39. R.H.S. \(= \tan(\alpha + \beta)\tan(\alpha - \beta)\)

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

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

    \(= \frac{\sin^2\alpha\cos^2\beta - \sin^2\beta\cos^2\alpha}{\cos^2\alpha\cos^2\beta - \sin^2\alpha\sin^2\beta}\)

    \(= \frac{\sin^2\alpha(1 - \sin^2\beta) - \sin^2\beta(1 - \sin^2\alpha)}{\cos^2\alpha(1 - \sin^2\beta) - \sin^2\beta(1 - \cos^2\alpha)}\)

    \(= \frac{\sin^2\alpha - \sin^2\beta}{\cos^2\alpha - \sin^2\beta} =\) R.H.S.

  40. L.H.S. \(= \tan^2\alpha -\tan^2\beta = \frac{\sin^2\alpha}{\cos^2\alpha} - \frac{\sin^2\beta}{\cos^2\beta}\)

    \(= \frac{\sin^2\alpha\cos^2\beta - \sin^2\beta\cos^2\alpha}{\cos^2\alpha\cos^2\beta}\)

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

    \(= \frac{\sin(\alpha + \beta)\sin(\alpha - \beta)}{\cos^2\alpha\cos^2\beta} =\) R.H.S.

  41. L.H.S. \(= \tan[(2n + 1)\pi + \theta] + \tan[(2n + 1)\pi - \theta]\)

    \(= \tan(\pi + \theta) + \tan(\pi - \theta)~[\because \tan 2n\pi = 0]\)

    \(= \tan\theta - \tan\theta = 0 =\) R.H.S.

  42. L.H.S. \(= \tan\left(\frac{\pi}{4} + \theta\right)\tan\left(\frac{3\pi}{4} + \theta\right) + 1\)

    \(= \tan\left(\frac{\pi}{4} + \theta\right)\tan\left[\pi - \left(\frac{\pi}{4} - \theta\right)\right] + 1\)

    \(= -\tan\left(\frac{\pi}{4} + \theta\right)\tan\left(\frac{\pi}{4} - \theta\right) + 1\)

    \(= -\tan\left(\frac{\pi}{4} + \theta\right)\tan\left[\left(\frac{\pi}{2} - \frac{pi}{4} - \theta\right)\right] + 1\)

    \(= -\tan\left(\frac{\pi}{4} + \theta\right)\cot\left(\frac{\pi}{4} + \theta\right) + 1 = -1 + 1 = 0 =\) R.H.S.

  43. R.H.S. \(= \frac{1 - pq}{\sqrt{(1 + p^2)(1 + q^2)}}\)

    Substituting for \(p\) and \(q,\) we get

    \(= \frac{1 - \tan\alpha\tan\beta}{\sqrt{(1 + \tan^2\alpha)(1 + \tan^2\beta)}}\)

    \(= \frac{\frac{\cos\alpha\cos\beta - \sin\alpha\sin\beta}{\cos\alpha\cos\beta}}{\sqrt{\sec\alpha\sec\beta}}\)

    \(= \cos(\alpha + \beta) =\) R.H.S.

  44. Given, \(\tan \beta = \frac{2\sin\alpha\sin\gamma}{\sin(\alpha + \gamma)}\)

    Inverting, we get

    \(2\cot\beta = \frac{\sin(\alpha + \gamma)}{\sin\alpha\sin\gamma} = \frac{\sin\alpha\cos\gamma + \sin\gamma\cos\alpha}{\sin\alpha\sin\gamma}\)

    \(= \cot \alpha + \cot \gamma\)

    Thus, \(\cot\alpha, \cot\beta, \cot\gamma\) are in A.P.

  45. \(\tan(\theta + \alpha - (\theta - \alpha)) = \tan2\alpha = \frac{\tan(\theta + \alpha) - \tan(\theta - \alpha)}{1 + \tan(\theta + \alpha)\tan(\theta - \alpha)}\)

    \(= \frac{b - a}{1 + ab}\)

  46. \(\tan\gamma = \tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\)

    \(a - b = \cot\alpha + \cot\beta - \tan\alpha - \tan\beta\)

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

    \(= \sin(\alpha + \beta)\left(\frac{\cos\alpha\cos\beta - \sin\alpha\sin\beta}{\sin\alpha\sin\beta\cos\alpha\cos\beta}\right)\)

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

    \(ab = (\tan\alpha + \tan\beta)(\cot\alpha + \cot\beta)\)

    \(= \frac{\sin^2(\alpha + \beta)}{\sin\alpha\sin\beta\cos\alpha\cos\beta}\)

    \(\frac{ab}{a - b} = \tan(\alpha + \beta) = \tan\gamma\)

  47. Given, \(A + B = 45^\circ \therefore \tan(A + B) = 1\)

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

    \(1 + \tan A + \tan B + \tan A \tan B = 2\)

    \((1 + \tan A)(1 + \tan B) = 2\)

  48. Given, \(\sin\alpha\sin\beta - \cos\alpha\cos\beta + 1 = 0\)

    \(\Rightarrow \cos\alpha\cos\beta - \sin\alpha\sin\beta = 1\)

    \(\Rightarrow \cos(\alpha + \beta) = 1\)

    \(\Rightarrow \sin(\alpha + \beta) = 0\)

    \(1 + \cot\alpha\tan\beta = 1 + \frac{\cos\alpha\sin\beta}{\sin\alpha\cos\beta}\)

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

    \(= \frac{\sin(\alpha + \beta)}{\sin\alpha\cos\beta} = \frac{0}{\sin\alpha\cos\beta} = 0\)

  49. \(\tan\beta = \frac{n\sin\alpha\cos\alpha}{1 - n\sin^2\alpha} = \frac{\frac{n\sin\alpha\cos\alpha}{\cos^2\alpha}}{\frac{1}{\cos^2\alpha} - n\frac{\sin^2\alpha}{\cos^2\alpha}}\)

    \(= \frac{n\tan\alpha}{\sec^2\alpha - n\tan^2\alpha} = \frac{n\tan\alpha}{1 + (1 - n)\tan^2\alpha}\)

    Now, \(\tan(\alpha - \beta) = \frac{\tan\alpha - \frac{n\tan\alpha}{1 + (1 - n)\tan^2\alpha}}{1 - \tan\alpha\frac{n\tan\alpha}{1 + (1 - n)\tan^2\alpha}}\)

    \(= \frac{\tan\alpha + (1 - n)\tan^3\alpha - n\tan\alpha}{1 + (1 - n)\tan^2\alpha + n\tan^2\alpha}\)

    \(= \frac{(1 - n)\tan\alpha + (1 - n)tan^3\alpha}{1 + \tan^2\alpha}\)

    \(= \frac{(1 - n)\tan\alpha(1 + \tan^2\alpha)}{ 1 + \tan^2\alpha} = (1 - n)\tan\alpha\)

  50. Given, \(\cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta) = -\frac{3}{2}\)

    \(3 + 2\cos(\beta - \gamma) + 2\cos(\gamma - \alpha) + 2\cos(\alpha - \beta) = 0\)

    \(3 + 2(\cos\beta\cos\gamma + \sin\beta\sin\gamma) + 2(\cos\gamma\cos\alpha + \sin\gamma\sin\alpha) + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) = 0\)

    \((\cos^2\alpha + \sin^2\alpha) + (\cos^2\beta + \sin^2\beta) + (\cos^2\gamma + \sin^2\gamma) + 2(\cos\beta\cos\gamma + \sin\beta\sin\gamma) + 2(\cos\gamma\cos\alpha + \sin\gamma\sin\alpha) + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) = 0\)

    \((\cos\alpha + \cos\beta + \cos\gamma)^2 + (\sin\alpha + \sin\beta + \sin\gamma^2) = 0\)

    \(\cos\alpha + \cos\beta + \cos\gamma = \sin\alpha + \sin\beta + \sin\gamma = 0\)

  51. \(\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\)

    \(= \frac{\frac{m}{m + 1} + \frac{1}{2m + 1}}{1 - \frac{m}{m + 1}\frac{1}{2m + 1}}\)

    \(= \frac{2m^2 + m + m + 1}{2m^2 + 3m + 1 - n} = 1\)

    Thus, \(\alpha + beta = \frac{\pi}{4}\)

  52. Given \((\cot A - 1)(\cot B - 1) = 2\)

    \(\cot A\cot B - 1 - \cot A - \cot B = 0\)

    \(\cot A\cot B - 1 = \cot A + \cot B \Rightarrow \frac{\cot A\cot B - 1}{\cot A + \cot B} = 1\)

    \(\cot(A + B) = \cot 45^\circ\)

    Thus, \(A + B = 45^\circ\)

    which we have proved in reverse.

  53. Given, \(\tan\alpha - \tan\beta = x\) and \(\cot\beta - \cot\alpha = y,\) we have to prove that \(\cot(\alpha - \beta) = \frac{x + y}{xy}\)

    Let \(\cot(\alpha - \beta) = \frac{x + y}{xy} = \frac{\tan\alpha - \tan\beta + \cot\beta - \cot\alpha}{(\tan\alpha - \tan\beta)(\cot\beta - \cot\alpha)}\)

    \(\tan(\alpha - \beta) = \frac{(\tan\alpha - \tan\beta)(\cot\beta - \cot\alpha)}{\tan\alpha - \tan\beta + \cot\beta - \cot\alpha}\)

    \(= \frac{\tan\alpha - \tan\beta}{1 + \frac{\tan\alpha - \tan\beta}{\cot\beta - \cot \alpha}}\)

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

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

    Hence proved.

  54. Given \(\alpha + \beta + \gamma = 90^\circ = \frac{\pi}{2}\)

    \(\cot \alpha = \cot\left(\frac{\pi}{2} - (\beta + \gamma)\right) = \tan(\beta + \gamma)\)

    \(= \frac{\tan\beta + \tan\gamma}{1 - \tan\beta\tan\gamma}\)

  55. We have to prove that \(\cot \beta = 2\tan(\alpha - \beta)\)

    \(\frac{1}{\tan\beta} = 2\frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}\)

    \(1 + \tan\alpha\tan\beta = 2\tan\alpha\tan\beta - 2\tan^2\beta\)

    Dividing both sides by \(\tan\beta,\) we get

    \(\cot \beta + \tan\alpha = 2\tan\alpha - 2\tan\beta\)

    \(\cot\beta + 2\tan\beta = =\tan\alpha\)

    Hence proved.

  56. \(\sin A = \frac{a}{c}, \sin B = \frac{b}{c}, \cos A = \frac{b}{c}, \cos B = \frac{a}{c}\)

    \(\cosec(A - B) = \frac{1}{\sin(A - B)} = \frac{1}{\sin A\cos B - \cos A\sin B} = \frac{1}{\frac{a^2}{c^2} - \frac{b^2}{c^2}}\)

    \(= \frac{c^2}{a^2 - b^2} = \frac{a^2 + b^2}{a^2 - b^2}\)

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

  57. We have to prove that \(A + B = C\) i.e. \(\tan(A + B) = \tan C\)

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

    \(\frac{\frac{1}{\sqrt{ac}} + \sqrt{\frac{a}{c}}}{1 - \frac{1}{\sqrt{ac}}\sqrt{\frac{a}{c}}} = \sqrt{\frac{c}{a^3}}\)

    \(= \frac{\frac{1}{\sqrt{ac}} + \sqrt{\frac{a}{c}}}{1 - \frac{1}{c}}\)

    \(\frac{1 + a}{\sqrt{ac}}.\frac{c}{c - 1} = \sqrt{\frac{c}{a^3}}\)

    \(\frac{ac + c}{c - 1} = \frac{c}{a}\)

    \(a^2c + ac = c^2 - c\)

    \(a^2 + a + 1 = c\) which is given, hence proved.

  58. Given \(\frac{\tan(A - B)}{\tan A} + \frac{\sin^2C}{\sin^2A} = 1\)

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

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

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

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

    \(\Rightarrow \tan A\tan B= \tan^2 C\)

  59. Given, \(\sin\alpha\sin\beta - \cos\alpha\cos\beta = 1\)

    \(\cos(\alpha + \beta) = -1 \Rightarrow \alpha + \beta = (2n + 1)\pi\)

    \(\tan(\alpha + \beta) = 0 \Rightarrow \tan \alpha + \tan \beta = 0\)

  60. Given, \(\sin\theta = 3\sin(\theta + 2\alpha)\)

    \(\sin(\theta + \alpha - \alpha) = 3\sin(\theta + \alpha + \alpha)\)

    \(\sin(\theta + \alpha)\cos\alpha - \sin\alpha\cos(\theta + \alpha) = 3\sin(\theta + \alpha)\cos\alpha + 3\cos(\theta + \alpha)\sin\alpha\)

    \(2\sin(\theta + \alpha)\cos\alpha + 4\sin\alpha\cos(\theta + \alpha) = 0\)

    Dividingboth sides with \(2\cos(\theta + \alpha)\cos\alpha,\) we get

    \(\tan(\theta + \alpha) + 2\tan\alpha = 0\)

  61. Given, \(3\tan\theta\tan\phi = 1 \Rightarrow \cot\tehta\cot\phi = 3\)

    \(\frac{\cos\theta\cos\phi}{\sin\theta\sin\phi} = 3\)

    Applying componendo and dividendo

    \(\frac{\cos\theta\cos\phi + \sin\theta\sin\phi}{\cos\theta\cos\phi - \sin\theta\sin\phi} = \frac{3 + 1}{3 - 1}\)

    \(\cos(\theta - \phi) = 2\cos(\theta + \phi)\)

  62. Let \(z = \cos\theta + \sin\theta = \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos\theta + \frac{1}{\sqrt{2}}\sin\theta\right)\)

    \(= \sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right)\)

    \(= \sqrt{2}\cos 55^\circ\) which has positive sign.

  63. Let \(z = 5\cos\theta + 3\cos\left(\theta + \frac{\pi}{3}\right) + 3\)

    \(z = 5\cos\thete + \frac{3}{2}\cos\theta - \frac{3\sqrt{3}}{2}\sin\theta + 3\)

    \(= \frac{13}{2}\cos\theta - \frac{3\sqrt{3}}{2}\sin\theta + 3\)

    \(= 7\left(\frac{13}{14}\cos\theta - \frac{3\sqrt{3}}{14}\sin\theta\right) + 3\)

    Let \(\cos\alpha = \frac{13}{14}\) then \(\sin\alpha = \frac{3\sqrt{3}}{14}\)

    \(y = 7(\cos\alpha\cos\theta - \sin\alpha\sin\thete) + 3\)

    \(y = 7\cos(\theta + \alpha) + 3\)

    Now maximum and minimum values of \(\cos(\theta + \alpha)\) are \(1\) and \(-1.\) Thus, value of \(y\) will lie between \(4\) and \(10.\)

  64. Given, \(m\tan(\theta - 30^\circ) = n\tan(\theta + 120^\circ)\)

    \(\frac{\tan(\theta - 30^\circ)}{\tan(\theta + 120^\circ)} = \frac{n}{m}\)

    \(\frac{\sin(\theta - 30^\circ)\cos(\theta) + 120^\circ}{\cos(\theta - 30^\circ)\sin(\theta) + 120^\circ} = \frac{n}{m}\)

    Applying componendo and dividendo

    \(\frac{\sin[(\theta + 120^\circ) + (\theta - 30^\circ)]}{\sin[(\theta + 120^\circ) - (\theta - 30^\circ)} = \frac{m + n}{m - n}\)

    \(\frac{\sin(2\theta + 90^\circ)}{\sin150^\circ} = \frac{m + n}{m - n}\)

    \(\cos2\theta = \frac{m + n}{2(m - n)}\)

  65. Given, \(\frac{\tan\alpha}{\tan\beta} = \frac{x}{y}\)

    Applying componendo and dividendo

    \(\frac{\tan\alpha + \tan\beta}{\tan\alpha - \tan\beta} = \frac{x + y}{x - y}\)

    \(\frac{\sin(\alpha + \beta)}{\sin(\alpha - \beta)} = \frac{x + y}{x - y}\)

    \(\sin(\alpha - \beta) = \frac{x - y}{x + y}\sin\theta\)

  66. We have to find the maximum and minimul values of \(7\cos\theta + 24\sin\theta = y\) (let)

    \(= 25\left(\frac{7}{25}\cos\theta + \frac{24}{25}\sin\theta\right)\)

    If \(\cos\alpha = \frac{7}{25}\) then \(\sin\alpha = \frac{24}{75}\)

    \(y = 25\cos(\theta - \alpha)\)

    Thus, maximum and minimum values of \(y\) are \(25\) and \(-25.\)

  67. Given expression is \(\sin100^\circ - \sin10^\circ = \cos10^\circ - \sin10^\circ = y\) (let)

    \(y = \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos^10 - \frac{1}{\sqrt{2}\sin10^\circ}\right)\)

    \(= \sqrt{2}\cos(45^\circ + 10^\circ)\)

    Thus, the sign is positive.