# 9. Compound Angles Solutions¶

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