$\displaystyle \text{Question 1: If } \sin A = \frac{4}{5} \text{ and } \cos B = \frac{5}{13} \text{, where } 0 < A , B < \frac{\pi}{2}, \\ \\ \text{ find the values of the following: } \\ \\ \text{ (i) } \sin (A+B) \displaystyle \text{ (ii) } \cos (A+B) \displaystyle \text{ (iii) } \sin (A - B) \displaystyle \text{ (iv) } \cos(A-B)$

$\displaystyle \text{Given } \sin A = \frac{4}{5} \text{ and } \cos B = \frac{5}{13} \text{, where } 0 < A , B < \frac{\pi}{2}$

This means that A is in Quadrant I and B is in Quadrant I

$\displaystyle \text{Therefore, } \cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \Big( \frac{4}{5} \Big)^2 } = \sqrt{\frac{9}{25}} = \frac{3}{5}$

$\displaystyle \text{Similarly, } \sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \Big( \frac{5}{13} \Big)^2 } = \sqrt{\frac{144}{169}} = \frac{12}{13}$

$\displaystyle \text{ (i) } \sin ( A+B) = \sin A \cos B + \cos A \sin B \\ \\ = \frac{4}{5} \times \frac{5}{13} + \frac{3}{5} \times \frac{12}{13} = \frac{20+36}{65} = \frac{56}{65}$

$\displaystyle \text{ (ii) } \cos ( A+B) = \cos A \cos B - \sin A \sin B \\ \\ = \frac{3}{5} \times \frac{5}{13} - \frac{4}{5} \times \frac{12}{13} = \frac{15-48}{65} = \frac{-33}{65}$

$\displaystyle \text{ (iii) } \sin ( A-B) = \sin A \cos B - \cos A \sin B \\ \\ = \frac{4}{5} \times \frac{5}{13} - \frac{3}{5} \times \frac{12}{13} = \frac{20-36}{65} = \frac{-16}{65}$

$\displaystyle \text{ (iv) } \cos ( A-B) = \cos A \cos B + \sin A \sin B \\ \\ = \frac{3}{5} \times \frac{5}{13} + \frac{4}{5} \times \frac{12}{13} = \frac{15+48}{65} = \frac{63}{65}$

$\displaystyle \\$

Question 2:

$\displaystyle \text{a) If } \sin A = \frac{12}{13} \text{ and } \sin B = \frac{4}{5} \text{, where } \frac{\pi}{2} < A < \pi , 0 < B < \frac{\pi}{2}, \\ \\ \text{ find the values of the following: } \text{ (i) } \sin (A+B) \displaystyle \text{ (ii) } \cos (A+B)$

$\displaystyle \text{b) If } \sin A = \frac{3}{5} \text{ and } \cos B = - \frac{12}{13} \text{, where } A \\ \\ \text{ and } B \text{ both lie in Q II find the value of } \sin (A+B)$

$\displaystyle \text{a) Given } \sin A = \frac{12}{13} \text{ and } \sin B = \frac{4}{5} \text{, where } \frac{\pi}{2} < A < \pi , 0 < B < \frac{\pi}{2}$

This means that A is in Quadrant II and B is in Quadrant I

$\displaystyle \text{Therefore, } \cos A = -\sqrt{1 - \sin^2 A} = - \sqrt{1 - \Big( \frac{12}{13} \Big)^2 } = - \sqrt{\frac{25}{69}} = - \frac{5}{13}$

$\displaystyle \text{Similarly, } \cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \Big( \frac{4}{5} \Big)^2 } = \sqrt{\frac{9}{25}} = \frac{3}{5}$

$\displaystyle \text{ (i) } \sin ( A+B) = \sin A \cos B + \cos A \sin B \\ \\ = \frac{12}{13} \times \frac{3}{5} + \frac{-5}{13} \times \frac{4}{5} = \frac{36-20}{65} = \frac{-16}{65}$

$\displaystyle \text{ (ii) } \cos ( A+B) = \cos A \cos B - \sin A \sin B \\ \\ = \frac{-5}{13} \times \frac{3}{5} - \frac{12}{13} \times \frac{4}{5} = \frac{-15-48}{65} = \frac{-63}{65}$

$\displaystyle \text{b) Given } \sin A = \frac{3}{5} \text{ and } \cos B = - \frac{12}{13} \text{, where } A \text{ and } B \text{ both lie in Q II }$

$\displaystyle \text{Therefore, } \cos A = -\sqrt{1 - \sin^2 A} = - \sqrt{1 - \Big( \frac{3}{5} \Big)^2 } = - \sqrt{\frac{16}{25}} = - \frac{4}{5}$

$\displaystyle \text{Similarly, } \sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \Big( \frac{-12}{13} \Big)^2 } = \sqrt{\frac{25}{169}} = \frac{5}{13}$

$\displaystyle \sin ( A+B) = \sin A \cos B + \cos A \sin B \\ \\ = \frac{3}{5} \times \frac{-12}{13} + \frac{-4}{5} \times \frac{5}{13} = \frac{-36-20}{65} = \frac{-56}{65}$

$\displaystyle \\$

$\displaystyle \text{Question 3: If } \cos A = - \frac{24}{25} \text{ and } \cos B = \frac{3}{5} \text{, where } \pi < A < \frac{3\pi}{2} , \frac{3\pi}{2} < B < 2\pi, \text{ find the values of the following: } \\ \\ \text{ (i) } \sin (A+B) \displaystyle \text{ (ii) } \cos (A+B)$

$\displaystyle \text{Given } \cos A = - \frac{24}{25} \text{ and } \cos B = \frac{3}{5} \text{, where } \pi < A < \frac{3\pi}{2}$, $\displaystyle \frac{3\pi}{2} < B < 2\pi$

This means that A is in Quadrant III and B is in Quadrant IV

$\displaystyle \text{Therefore, } \sin A = -\sqrt{1 - \cos^2 A} = - \sqrt{1 - \Big( \frac{-24}{25} \Big)^2 } = - \sqrt{\frac{49}{625}} = - \frac{7}{25}$

$\displaystyle \text{Similarly, } \sin B = -\sqrt{1 - \cos^2 B} = - \sqrt{1 - \Big( \frac{3}{5} \Big)^2 } = - \sqrt{\frac{16}{25}} = - \frac{4}{5}$

$\displaystyle \text{ (i) } \sin ( A+B) = \sin A \cos B + \cos A \sin B \\ \\ = \frac{-7}{25} \times \frac{3}{5} + \frac{-24}{25} \times \frac{-4}{5} = \frac{-21+96}{125} = \frac{3}{5}$

$\displaystyle \text{ (ii) } \cos ( A+B) = \cos A \cos B - \sin A \sin B \\ \\ = \frac{-24}{25} \times \frac{3}{5} - \frac{-7}{25} \times \frac{-4}{5} = \frac{-72-28}{125} = \frac{-4}{5}$

$\displaystyle \\$

$\displaystyle \text{Question 4: If } \tan A = \frac{3}{4} \text{ and } \cos B = \frac{9}{41} \text{, where } \pi < A < \frac{3\pi}{2} \text{ and } 0 < B < \frac{\pi}{2} , \text{ find } \tan (A+B)$

$\displaystyle \text{Given } \tan A = \frac{3}{4} \text{ and } \cos B = \frac{9}{41} \text{, where } \pi < A < \frac{3\pi}{2} \text{ and } 0 < B < \frac{\pi}{2}$

This means that A is in Quadrant III and B is in Quadrant I

$\displaystyle \text{Therefore, } \sin B = -\sqrt{1 - \cos^2 B} = \sqrt{1 - \Big( \frac{9}{41} \Big)^2 } = \sqrt{\frac{1600}{1681}} = \frac{40}{41}$

$\displaystyle \therefore \tan B = \frac{\sin B}{\cos B} = \frac{\frac{40}{41}}{\frac{9}{41}} = \frac{40}{9}$

$\displaystyle \therefore \tan (A+B) = \frac{\tan A + \tan B}{1- \tan A \tan B} = \frac{\frac{3}{4} + \frac{40}{9}}{1- \frac{3}{4} \times \frac{40}{9}} = \frac{27+ 160}{36-120} = - \frac{187}{84}$

$\displaystyle \\$

$\displaystyle \text{Question 5: If } \sin A = \frac{1}{2} \text{ and } \cos B = \frac{12}{13} \text{, where } \frac{\pi}{2} < A < \pi \text{ and } \frac{3\pi}{2} < B < 2\pi , \text{ find } \tan (A-B)$

$\displaystyle \text{Given } \sin A = \frac{1}{2} \text{ and } \cos B = \frac{12}{13} \text{, where } \frac{\pi}{2} < A < \pi \text{ and } \frac{3\pi}{2} < B < 2\pi$

This means that A is in Quadrant II and B is in Quadrant IV

$\displaystyle \text{Therefore, } \cos A = -\sqrt{1 - \sin^2 A} = - \sqrt{1 - \Big( \frac{1}{2} \Big)^2 } = - \sqrt{\frac{3}{4}} = - \frac{\sqrt{3}}{2}$

$\displaystyle \text{Similarly, } \sin B = -\sqrt{1 - \cos^2 B} = - \sqrt{1 - \Big( \frac{12}{13} \Big)^2 } = - \sqrt{\frac{25}{169}} = - \frac{5}{13}$

$\displaystyle \therefore \tan A = \frac{\sin A}{\cos A} = \frac{\frac{1}{2}}{\frac{-\sqrt{3}}{2}} = - \frac{1}{\sqrt{3}}$

$\displaystyle \text{Similarly, } \tan B = \frac{\sin B}{\cos B} = \frac{\frac{-5}{13}}{\frac{12}{13}} = - \frac{5}{12}$

$\displaystyle \therefore \tan (A-B) = \frac{\tan A - \tan B}{1+ \tan A \tan B} = \frac{\frac{-1}{\sqrt{3}} - \frac{-5}{12}}{1+ \frac{-1}{\sqrt{3}} \times \frac{-5}{12}} = \frac{-12+ 5\sqrt{3}}{12\sqrt{3}+5} = \frac{5\sqrt{3} - 12}{12\sqrt{3}+5}$

$\displaystyle \\$

$\displaystyle \text{Question 6: If } \sin A = \frac{1}{2} \text{ and } \cos B = \frac{\sqrt{3}}{2} \text{, where } \frac{\pi}{2} < A < \pi \text{ and } 0 < B < \frac{\pi}{2},$,

Find the following: $\displaystyle \text{ (i) } \tan (A+B) \displaystyle \text{ (ii) } \tan (A-B)$

$\displaystyle \text{Given } \sin A = \frac{1}{2} \text{ and } \cos B = \frac{\sqrt{3}}{2} \text{, where } \frac{\pi}{2} < A < \pi \text{ and } 0 < B < \frac{\pi}{2}$

This means that A is in Quadrant II and B is in Quadrant I

$\displaystyle \text{Therefore, } \cos A = -\sqrt{1 - \sin^2 A} = - \sqrt{1 - \Big( \frac{1}{2} \Big)^2 } = - \sqrt{\frac{3}{4}} = - \frac{\sqrt{3}}{2}$

$\displaystyle \text{Similarly, } \sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \Big( \frac{\sqrt{3}}{2} \Big)^2 } = \sqrt{\frac{1}{4}} = \frac{1}{2}$

$\displaystyle \therefore \tan A = \frac{\sin A}{\cos A} = \frac{\frac{1}{2}}{\frac{-\sqrt{3}}{2}} = - \frac{1}{\sqrt{3}}$

$\displaystyle \text{Similarly, } \tan B = \frac{\sin B}{\cos B} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$

$\displaystyle \text{ (i) } \tan (A+B) = \frac{\tan A + \tan B}{1- \tan A \tan B} = \frac{\frac{-1}{\sqrt{3}}+ \frac{1}{\sqrt{3}}}{1 - \frac{-1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}} = \frac{-1+1}{1+\frac{1}{3}} = 0$

$\displaystyle \text{ (ii) } \tan (A-B) = \frac{\tan A - \tan B}{1+ \tan A \tan B} = \frac{\frac{-1}{\sqrt{3}}- \frac{1}{\sqrt{3}}}{1 + \frac{-1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}} = \frac{-2\sqrt{3}}{2} = -\sqrt{3}$

$\displaystyle \\$

Question 7: Evaluate the following:

$\displaystyle \text{ (i) } \sin 78^{\circ} \cos 18^{\circ} - \cos 78^{\circ} \sin 18^{\circ} \displaystyle \text{ (ii) } \cos 47^{\circ} \cos 13^{\circ} - \sin 47^{\circ} \sin 13^{\circ}$

$\displaystyle \text{ (iii) } \sin 36^{\circ} \cos 9^{\circ} + \cos 36^{\circ} \sin 9^{\circ}$ $\displaystyle \text{ (iv) } \cos 80^{\circ} \cos 20^{\circ} + \sin 80^{\circ} \sin 20^{\circ}$

$\displaystyle \text{ (i) } \sin 78^{\circ} \cos 18^{\circ} - \cos 78^{\circ} \sin 18^{\circ} = \sin (78^{\circ}-18^{\circ}) = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$

$\displaystyle \text{ (ii) } \cos 47^{\circ} \cos 13^{\circ} - \sin 47^{\circ} \sin 13^{\circ} = \cos (47^{\circ}+13^{\circ}) = \cos 60^{\circ} = \frac{1}{2}$

$\displaystyle \text{ (iii) } \sin 36^{\circ} \cos 9^{\circ} + \cos 36^{\circ} \sin 9^{\circ} = \sin ( 36^{\circ} + 9^{\circ}) = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$

$\displaystyle \text{ (iv) } \cos 80^{\circ} \cos 20^{\circ} + \sin 80^{\circ} \sin 20^{\circ} = \cos (80^{\circ}-20^{\circ}) = \cos 60^{\circ} = \frac{1}{2}$

$\displaystyle \\$

Question 8: $\displaystyle \text{If } \cos A = - \frac{12}{13} \text{ and } \cot B = \frac{24}{7}$, where A lies in the second quadrant and B in the third quadrant, find the values of the following:

$\displaystyle \text{ (i) } \sin (A+B) \displaystyle \text{ (ii) } \cos (A+B) \displaystyle \text{ (iii) } \tan (A + B)$

$\displaystyle \text{Given } \cos A = - \frac{12}{13} \text{ and } \cot B = \frac{24}{7},$

where A lies in the second quadrant and B in the third quadrant

$\displaystyle \text{Therefore, } \sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \Big( \frac{-12}{13} \Big)^2 } = \sqrt{\frac{25}{169}} = \frac{\sqrt{5}}{13}$

$\displaystyle \sin B = - \frac{1}{\sqrt{1+ \cot^2 B}} = - \frac{1}{\sqrt{1+ \Big( \frac{24}{7} \Big)^2}} = - \frac{1}{\sqrt{\frac{625}{49}}} = - \frac{7}{25}$

$\displaystyle \text{ (i) } \sin ( A+B) = \sin A \cos B + \cos A \sin B \\ \\ = \frac{5}{13} \times \frac{-24}{25} + \frac{-12}{13} \times \frac{-7}{25} = \frac{-120+84}{325} = \frac{-36}{325}$

$\displaystyle \text{ (ii) } \cos ( A+B) = \cos A \cos B - \sin A \sin B \\ \\ = \frac{-12}{13} \times \frac{-24}{25} - \frac{5}{13} \times \frac{-7}{25} = \frac{288+35}{325} = \frac{323}{325}$

$\displaystyle \tan ( A+B) = \frac{\sin (A+B)}{\cos (A+B) } = \frac{\frac{-36}{325}}{\frac{323}{325} } = \frac{-36}{323}$

$\displaystyle \\$

$\displaystyle \text{Question 9: Prove that } \cos \frac{7\pi}{12} + \cos \frac{\pi}{12} = \sin \frac{5\pi}{12} - \sin \frac{\pi}{12}$

$\displaystyle \text{LHS } = \cos \frac{7\pi}{12} + \cos \frac{\pi}{12}$

$\displaystyle = \cos \Big( \frac{\pi}{2} + \frac{\pi}{12} \Big) + \cos \Big( \frac{\pi}{2} - \frac{5\pi}{12} \Big)$

$\displaystyle = \sin \frac{5\pi}{12} - \sin \frac{\pi}{12}$

$\displaystyle =$ RHS. Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 10: Prove that } \frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\sin (A+B)}{\sin (A-B)}$

$\displaystyle \text{LHS } = \frac{\tan A + \tan B}{\tan A - \tan B}$

$\displaystyle = \frac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A} - \frac{\sin B}{\cos B}}$

$\displaystyle = \frac{\sin A \cos B + \cos A \sin B}{\sin A \cos B - \cos A \sin B}$

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

$\displaystyle =$ RHS. Hence proved.

$\displaystyle \\$

Question 11:

$\displaystyle \text{ (i) } \frac{\cos 11^{\circ} + \sin 11^{\circ}}{\cos 11^{\circ} - \sin 11^{\circ}} = \tan 56^{\circ}$

$\displaystyle \text{ (ii) } \frac{\cos 9^{\circ} + \sin 9^{\circ}}{\cos 9^{\circ} - \sin 9^{\circ}} = \tan 54^{\circ}$

$\displaystyle \text{ (iii) } \frac{\cos 8^{\circ} - \sin 8^{\circ}}{\cos 8^{\circ} + \sin 8^{\circ}} = \tan 37^{\circ}$

$\displaystyle \text{ (i) } \frac{\cos 11^{\circ} + \sin 11^{\circ}}{\cos 11^{\circ} - \sin 11^{\circ}} \\ \\ = \frac{1+\tan 11^{\circ}}{1- \tan 11^{\circ}} = \frac{\tan 45^{\circ} + \tan 11^{\circ}}{1 - \tan 45^{\circ} \tan 11^{\circ}} = \tan (45^{\circ} + 11^{\circ}) = \tan 56^{\circ}$

$\displaystyle \text{ (ii) } \frac{\cos 9^{\circ} + \sin 9^{\circ}}{\cos 9^{\circ} - \sin 9^{\circ}} \\ \\ = \frac{1+\tan 9^{\circ}}{1- \tan 9^{\circ}} = \frac{\tan 45^{\circ} + \tan 9^{\circ}}{1 - \tan 45^{\circ} \tan 9^{\circ}} = \tan (45^{\circ} + 9^{\circ}) = \tan 54^{\circ}$

$\displaystyle \text{ (iii) } \frac{\cos 8^{\circ} - \sin 8^{\circ}}{\cos 8^{\circ} +\sin 8^{\circ}} \\ \\ = \frac{1-\tan 8^{\circ}}{1+ \tan 8^{\circ}} = \frac{\tan 45^{\circ} - \tan 8^{\circ}}{1 + \tan 45^{\circ} \tan 8^{\circ}} = \tan (45^{\circ} - 8^{\circ}) = \tan 37^{\circ}$

$\displaystyle \\$

Question 12:

$\displaystyle \text{ (i) } \sin \Big( \frac{\pi}{3} - x \Big) \cos \Big( \frac{\pi}{6} + x \Big) + \cos \Big( \frac{\pi}{3} - x \Big) \sin \Big( \frac{\pi}{6} + x \Big) = 1$

$\displaystyle \text{ (ii) } \sin \Big( \frac{4\pi}{9} + 7 \Big) \cos \Big( \frac{\pi}{9} + 7 \Big) - \cos \Big( \frac{4\pi}{9} +7 \Big) \sin \Big( \frac{\pi}{9} + 7 \Big) = \frac{\sqrt{3}}{2}$

$\displaystyle \text{ (iii) } \sin \Big( \frac{3\pi}{8} - 5 \Big) \cos \Big( \frac{\pi}{8} + 5 \Big) + \cos \Big( \frac{3\pi}{8} -5 \Big) \sin \Big( \frac{\pi}{9} + 5 \Big) = 1$

$\displaystyle \text{(i) } \text{LHS } = \sin \Big( \frac{\pi}{3} - x \Big) \cos \Big( \frac{\pi}{6} + x \Big) + \cos \Big( \frac{\pi}{3} - x \Big) \sin \Big( \frac{\pi}{6} + x \Big)$

Note: $\displaystyle \sin (A + B) = \sin A \cos B + \cos A + \sin B$

$\displaystyle = \sin \Big[ \Big( \frac{\pi}{3} - x \Big) + \Big( \frac{\pi}{6} + x \Big) \Big]$

$\displaystyle = \sin \Big[ \frac{\pi}{3} + \frac{\pi}{6} \Big]$

$\displaystyle = \sin \frac{\pi}{2}$

$\displaystyle = 1 =$ RHS. Hence proved.

$\displaystyle \text{(ii) } \text{LHS } = \sin \Big( \frac{4\pi}{9} + 7 \Big) \cos \Big( \frac{\pi}{9} + 7 \Big) - \cos \Big( \frac{4\pi}{9} +7 \Big) \sin \Big( \frac{\pi}{9} + 7 \Big)$

Note: $\displaystyle \sin (A - B) = \sin A \cos B - \cos A + \sin B$

$\displaystyle = \sin \Big[ \Big( \frac{4\pi}{9} + 7 \Big) - \Big( \frac{\pi}{9} + 7 \Big) \Big]$

$\displaystyle = \sin \Big[ \frac{4\pi}{9} - \frac{\pi}{9} \Big]$

$\displaystyle = \sin \frac{3\pi}{9}$

$\displaystyle = \frac{\sqrt{3}}{2}$ RHS. Hence proved.

$\displaystyle \text{(iii) } \text{LHS } = \sin \Big( \frac{3\pi}{8} - 5 \Big) \cos \Big( \frac{\pi}{8} + 5 \Big) + \cos \Big( \frac{3\pi}{8} -5 \Big) \sin \Big( \frac{\pi}{9} + 5 \Big)$

Note: $\displaystyle \sin (A + B) = \sin A \cos B + \cos A + \sin B$

$\displaystyle = \sin \Big[ \Big( \frac{3\pi}{8} - 5 \Big) + \Big( \frac{\pi}{8} + 5 \Big) \Big]$

$\displaystyle = \sin \Big[ \frac{3\pi}{8} + \frac{\pi}{8} \Big]$

$\displaystyle = \sin \frac{\pi}{2}$

$\displaystyle = 1 =$ RHS. Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 13: Prove that: } \frac{\tan 69^{\circ} + \tan 66^{\circ}}{1 - \tan 69^{\circ} \tan 66^{\circ}} =- 1$

$\displaystyle \text{LHS } = \frac{\tan 69^{\circ} + \tan 66^{\circ}}{1 - \tan 69^{\circ} \tan 66^{\circ}}$

$\displaystyle \text{Note: Since } \tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

$\displaystyle = \tan (69^{\circ}+66^{\circ}) = \tan 135^{\circ} - \cot 45^{\circ} = -1 = \text{RHS. Hence proved.}$

$\displaystyle \\$

Question 14:

$\displaystyle \text{ (i) } \text{If } \tan A = \frac{5}{6} \text{ and } \tan B = \frac{1}{11} , \text{ prove that } A + B = \frac{\pi}{4}$

$\displaystyle \text{ (ii) } \text{If } \tan A = \frac{m}{m-1} \text{ and } \tan B = \frac{1}{2m-1 } , \text{ prove that } A - B = \frac{\pi}{4}$

$\displaystyle \text{ (i) } \tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

$\displaystyle = \frac{\frac{5}{6} + \frac{1}{11}}{1 - \frac{5}{6} . \frac{1}{11}} = \frac{55 + 6}{66-5} = \frac{61}{61}$

$\displaystyle = 1 = \tan \frac{\pi}{4}$

$\displaystyle \therefore A+B = \frac{\pi}{4}$

$\displaystyle \text{ (ii) } \tan (A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

$\displaystyle = \frac{\frac{m}{m-1} - \frac{1}{2m-1 }}{1 + \frac{m}{m-1} . \frac{1}{2m-1 }}$

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

$\displaystyle = \frac{2m^2 - m - m + 1}{2m^2 - m - 2m + 1 + m}$

$\displaystyle = \frac{2m^2 - 2m + 1}{2m^2 - 2m + 1}$

$\displaystyle = 1 = \tan \frac{\pi}{4}$

$\displaystyle \therefore A-B = \frac{\pi}{4}$

$\displaystyle \\$

Question 15: Prove that

$\displaystyle \text{ (i) } \cos^2 \frac{\pi}{4} - \sin^2 \frac{\pi}{12} = \frac{\sqrt{3}}{4}$

$\displaystyle \text{ (ii) } \sin^2 (n+1) A - \sin^2 nA = \sin(2n+1) A \sin A$

$\displaystyle \text{i) LHS } = \cos^2 \frac{\pi}{4} - \sin^2 \frac{\pi}{12}$

$\displaystyle = \Big( \frac{1}{\sqrt{2}} \Big)^2 - \sin^2 \frac{\pi}{12}$

$\displaystyle \text{Lets calculate } \sin^2 \frac{\pi}{12} = \sin^2 15^{\circ}$

$\displaystyle \sin 15^{\circ} = \sin ( 45^{\circ}- 30^{\circ})$

$\displaystyle = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ}$

$\displaystyle = \Big( \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} . \frac{1}{2} \Big)$

$\displaystyle = \frac{\sqrt{3} -1}{8}$

Substituting it back

$\displaystyle = \frac{1}{2} - \Big( \frac{\sqrt{3} -1}{8} \Big)$

$\displaystyle = \frac{1}{2} - \Big( \frac{(3 +1 - 2 \sqrt{3})^2}{8} \Big)$

$\displaystyle = \frac{1}{2} - \Big( \frac{4 - 2 \sqrt{3}}{8} \Big)$

$\displaystyle = \frac{1}{2} - \Big( \frac{2 - \sqrt{3}}{4} \Big)$

$\displaystyle = \frac{2 - 2 + \sqrt{3}}{4}$

$\displaystyle = \frac{\sqrt{3}}{4}$

$\displaystyle = \text{RHS. Hence proved.}$

$\displaystyle \text{(ii) } \text{LHS } = \sin^2 (n+1) A - \sin^2 nA$

Since: $\displaystyle \sin^2 A - \sin^2 B = \sin (A+B) \sin(A-B)$

$\displaystyle = \sin [ (n+1)A + nA ] \sin [(n+1)A - nA ]$

$\displaystyle = \sin (2nA + A) \sin A$

$\displaystyle = \sin (2n+1)A \sin A$

$\displaystyle = \text{RHS. Hence proved.}$

$\displaystyle \\$

Question 16: Prove that:

$\displaystyle \text{ (i) } \frac{\sin (A+B) + \sin (A-B)}{\cos (A+B) + \cos (A-B)} = \tan A$

$\displaystyle \text{ (ii) } \frac{\sin (A-B)}{\cos A \cos B} + \frac{\sin (B-C)}{\cos B \cos C} + \frac{\sin (C-A)}{\cos C \cos A} = 0$

$\displaystyle \text{ (iii) } \frac{\sin (A-B)}{\sin A \sin B} + \frac{\sin (B-C)}{\sin B \sin C} + \frac{\sin (C-A)}{\sin C \sin A} = 0$

$\displaystyle \text{ (iv) } \sin^2 B= \sin^2 A + \sin^2 (A-B) - 2 \sin A \cos B \sin (A-B)$

$\displaystyle \text{ (v) } \cos^2 A + \cos^2 B - 2 \cos A \cos B \cos (A+ B) = \sin^2 (A+B)$

$\displaystyle \text{(vi) } \frac{\tan (A+B)}{\cot(A-B)} = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}$

$\displaystyle \text{(i) } \text{LHS } = \frac{\sin (A+B) + \sin (A-B)}{\cos (A+B) + \cos (A-B)}$

$\displaystyle = \frac{2 \sin A \cos B}{2 \cos A \cos B} = \tan A = \text{RHS. Hence proved.}$

$\displaystyle \text{(ii) } \text{LHS } = \frac{\sin (A-B)}{\cos A \cos B} + \frac{\sin (B-C)}{\cos B \cos C} + \frac{\sin (C-A)}{\cos C \cos A}$

$\displaystyle = \frac{\sin A \cos B - \cos A \sin B}{\cos A \cos B} + \frac{\sin B \cos C - \cos B \sin C}{\cos B \cos C} + \frac{\sin C \cos A - \cos C \sin A}{\cos C \cos A}$

$\displaystyle = \tan A - \tan B + \tan B - \tan C + \tan C - \tan A$

$\displaystyle = 0 = \text{RHS. Hence proved.}$

$\displaystyle \text{(iii) } \text{LHS } = \frac{\sin (A-B)}{\sin A \sin B} + \frac{\sin (B-C)}{\sin B \sin C} + \frac{\sin (C-A)}{\sin C \sin A}$

$\displaystyle = \frac{\sin A \cos B - \cos A \sin B}{\sin A \sin B} + \frac{\sin B \cos C - \cos B \sin C}{\sin B \sin C} + \frac{\sin C \cos A - \cos C \sin A}{\sin C \sin A}$

$\displaystyle = \cot B - \cot A + \cot C - \cot B + \cot A - \cot C$

$\displaystyle = 0 = \text{RHS. Hence proved.}$

$\displaystyle \text{iv) RHS }= \sin^2 A + \sin^2 (A-B) - 2 \sin A \cos B \sin (A-B)$

$\displaystyle = \sin^2 A + \sin (A_B) [ \sin (A-B) - 2 \sin A \cos B ]$

$\displaystyle = \sin^2 A + \sin (A - B) [ \sin A \cos B - \cos A \sin B - 2 \sin A \cos B ]$

$\displaystyle = \sin^2 A - \sin (A - B) [ \sin A \cos B + \cos A \sin B ]$

$\displaystyle = \sin^2 A - \sin (A - B) \sin ( A + B)$

$\displaystyle = \sin^2 - ( \sin^2 A - \sin^2 B)$

$\displaystyle = \sin^2 B = \text{LHS. Hence proved.}$

$\displaystyle \text{v) LHS } = \cos^2 A + \cos^2 B - 2 \cos A \cos B \cos (A+ B)$

$\displaystyle = \cos^2 + [1 - \sin^2 B ] - 2 \cos A \cos B \cos (A+B)$

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

$\displaystyle = \cos (A+B) \cos (A - B) - 2 \cos A \cos B \cos ( A + B) + 1$

$\displaystyle = \cos ( A+B) [ \cos (A - B) - 2 \cos A \cos B ] + 1$

$\displaystyle = \cos ( A + B) [ \cos A \cos B + \sin A \sin B - 2 \cos A \cos B ] + 1$

$\displaystyle = \cos (A+B) [ - \cos A \cos B + \sin A \sin B ] + 1$

$\displaystyle = - \cos (A + B) [\cos A \cos B - \sin A \sin B ] + 1$

$\displaystyle = - \cos (A + B) \cos ( A + B) + 1$

$\displaystyle = - \cos^2 ( A + B) + 1 = \sin^2 ( A + B) = \text{RHS. Hence proved.}$

$\displaystyle \text{(vi) } \text{LHS } = \frac{\tan (A+B)}{\cot(A-B)}$

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

$\displaystyle = \tan ( A + B ) \tan ( A - B)$

$\displaystyle = \Big[ \frac{\tan A + \tan B}{1 - \tan A \tan B} \Big] + \Big[ \frac{\tan A - \tan B}{1 + \tan A \tan B} \Big]$

$\displaystyle = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B} = \text{RHS. Hence proved.}$

$\displaystyle \\$

Question 17: Prove that:

$\displaystyle \text{ (i) } \tan 8x - \tan 6x - \tan 2x = \tan 8x \tan 6x \tan 2x$

$\displaystyle \text{ (ii) } \tan \frac{\pi}{12} + \tan \frac{\pi}{6} + \tan \frac{\pi}{12} \tan \frac{\pi}{6} = 1$

$\displaystyle \text{ (iii) } \tan 36^{\circ} + \tan 9^{\circ} + \tan 36^{\circ} \tan 9^{\circ} = 1$

$\displaystyle \text{ (iv) } \tan 13x - \tan 9x - \tan 4x = \tan 13x \tan 9x \tan 4x$

$\displaystyle \text{ (i) } 8x = 6x + 2x$

$\displaystyle \Rightarrow \tan 8x = \tan (6x + 2x)$

$\displaystyle \Rightarrow \tan 8x = \frac{\tan 6x + \tan 2x}{1- \tan 6x \tan 2x}$

$\displaystyle \Rightarrow \tan 8x ( 1- \tan 6x \tan 2x ) = \tan 6x + \tan 2x$

$\displaystyle \Rightarrow \tan 8x - \tan 8x \tan 6x \tan 2x = \tan 6x + \tan 2x$

$\displaystyle \Rightarrow \tan 8x - \tan 6x - \tan 2x = \tan 8x \tan 6x \tan 2x$

$\displaystyle \text{ (ii) } \frac{\pi}{4} = \frac{\pi}{6} + \frac{\pi}{12}$

$\displaystyle \Rightarrow \tan \frac{\pi}{4} = \tan (\frac{\pi}{6} + \frac{\pi}{12})$

$\displaystyle \Rightarrow \tan \frac{\pi}{4} = \frac{\tan \frac{\pi}{6}+ \tan \frac{\pi}{12}}{1- \tan \frac{\pi}{6} \tan \frac{\pi}{12}}$

$\displaystyle \Rightarrow 1 ( 1- \tan \frac{\pi}{6} \tan \frac{\pi}{12} ) = \tan \frac{\pi}{6} + \tan \frac{\pi}{12}$

$\displaystyle \Rightarrow 1 - \tan \frac{\pi}{6} \tan \frac{\pi}{12} = \tan \frac{\pi}{6} + \tan \frac{\pi}{12}$

$\displaystyle \Rightarrow 1 = \tan \frac{\pi}{6} + \tan \frac{\pi}{12} + \tan \frac{\pi}{6} \tan \frac{\pi}{12}$

$\displaystyle \text{ (iii) } 45^{\circ} = 36^{\circ} + 9^{\circ}$

$\displaystyle \Rightarrow \tan 45^{\circ} = \tan (36^{\circ} + 9^{\circ})$

$\displaystyle \Rightarrow \tan 45^{\circ} = \frac{\tan 36^{\circ} + \tan 9^{\circ}}{1- \tan 36^{\circ} \tan 9^{\circ}}$

$\displaystyle \Rightarrow 1 ( 1- \tan 36^{\circ} \tan 9^{\circ} ) = \tan 36^{\circ} + \tan 9^{\circ}$

$\displaystyle \Rightarrow 1 - \tan 36^{\circ} \tan 9^{\circ} = \tan 36^{\circ} + \tan 9^{\circ}$

$\displaystyle \Rightarrow 1 - \tan 36^{\circ} - \tan 9^{\circ} = \tan 36^{\circ} \tan 9^{\circ}$

$\displaystyle \text{ (iv) } 13x = 9x + 4x$

$\displaystyle \Rightarrow \tan 13x = \tan (9x + 4x)$

$\displaystyle \Rightarrow \tan 13x = \frac{\tan 9x + \tan 4x}{1- \tan 9x \tan 4x}$

$\displaystyle \Rightarrow \tan 13x ( 1- \tan 9x \tan 4x ) = \tan 9x + \tan 4x$

$\displaystyle \Rightarrow \tan 13x - \tan 13x \tan 9x \tan 4x = \tan 9x + \tan 4x$

$\displaystyle \Rightarrow \tan 13x - \tan 9x - \tan 4x = \tan 13x \tan 9x \tan 4x$

$\displaystyle \\$

$\displaystyle \text{Question 18: Prove that } \frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x} = \tan 3x \tan x$

RHS $\displaystyle = \tan 3x \tan x$

$\displaystyle = \tan (2x+x) \tan (2x-x)$

$\displaystyle = [ \frac{\tan 2x + \tan x}{1- \tan 2x \tan x} ] [ \frac{\tan 2x - \tan x}{1+ \tan 2x \tan x} ]$

$\displaystyle = \frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x}$

$\displaystyle =$ LHS. Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 19: } \frac{\sin(x+y)}{\sin(x-y)} = \frac{a+b}{a-b}, \text{ show that } \frac{\tan x}{\tan y} = \frac{a}{b}$

$\displaystyle \frac{\sin(x+y)}{\sin(x-y)} = \frac{a+b}{a-b}$

$\displaystyle \Rightarrow \frac{\sin x \cos y + \cos x \sin y}{\sin x \cos y - \cos x \sin y} = \frac{a+b}{a-b}$

Using Componendo and Dividendo

$\displaystyle \Rightarrow \frac{\sin x \cos y + \cos x \sin y + \sin x \cos y - \cos x \sin y}{\sin x \cos y + \cos x \sin y - \sin x \cos y + \cos x \sin y} = \frac{a+b + a - b}{a + b - a+b}$

$\displaystyle \Rightarrow \frac{2\sin x \cos y}{2\cos x \sin y} = \frac{2a}{2b}$

$\displaystyle \Rightarrow \frac{\sin x \cos y}{\cos x \sin y} = \frac{a}{b}$

$\displaystyle \Rightarrow \frac{\tan x}{\tan y} = \frac{a}{b}$

$\displaystyle \\$

$\displaystyle \text{Question 20: If } \tan A = x \tan B , \text{ prove that } \frac{\sin(A-B)}{\sin(A+B)} = \frac{x-1}{x+1}$

$\displaystyle \tan A = x \tan B$

$\displaystyle \frac{\sin A}{\cos A} = x \frac{\sin A}{\cos A}$

$\displaystyle \sin A \cos B = x \sin B \cos A$

$\displaystyle \text{Now } \frac{\sin (A-B)}{\sin (A+B)} = \frac{\sin A \cos B - \cos A \sin B}{\sin A \cos B + \cos A \sin B}$

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

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

$\displaystyle \frac{\sin (A-B)}{\sin (A+B)} = \frac{(x -1) }{(x - 1 ) }$

Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 21: If } \tan (A+B) = x \text{ and } \tan (A-B) = y$, find the value of $\displaystyle \tan 2A \text{ and } \tan 2B$

$\displaystyle \text{Given } \tan (A+B) = x \text{ and } \tan (A-B) = y$

$\displaystyle \tan 2A = \tan [ (A+B) + (A -B) ]$

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

$\displaystyle = \frac{x+y}{1-xy}$

Similarly,

$\displaystyle \tan 2B = \tan [ (A+B) - (A -B) ]$

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

$\displaystyle = \frac{x-y}{1+xy}$

$\displaystyle \\$

$\displaystyle \text{Question 22: If } \cos A + \sin B = m \text{ and } \sin A + \cos B = n$, prove that $\displaystyle 2 \sin (A+B) = m^2 + n^2 -2$

$\displaystyle \text{Given } \cos A + \sin B = m \text{ and } \sin A + \cos B = n$

$\displaystyle m^2 + n^2 - 2$

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

$\displaystyle = \cos^2 A + \sin^2 B + 2 \cos A \sin B + \sin^2 A + \cos^2 B + 2 \sin A \cos B - 2$

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

$\displaystyle = 2 + + 2 (\cos A \sin B + \sin A \cos B) - 2$

$\displaystyle = 2 (\cos A \sin B + \sin A \cos B)$

$\displaystyle = 2 \sin ( A + B)$ Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 23: If } \tan A + \tan B = a \text{ and } \cot A + \cot B = b$, prove that: $\displaystyle \cot (A+B) = \frac{1}{a} - \frac{1}{b}$

$\displaystyle \text{Given } \tan A + \tan B = a \text{ and } \cot A + \cot B = b$

$\displaystyle \cot A + \cot B = b$

$\displaystyle \Rightarrow \frac{1}{\tan A} + \frac{1}{\tan B} = b$

$\displaystyle \Rightarrow \frac{\tan A + \tan B}{1 - \tan A \tan B} = b$

$\displaystyle \Rightarrow \tan A \tan B = \frac{a}{b}$

$\displaystyle \because \cot ( A + B) = \frac{1}{\tan ( A + B)}$

$\displaystyle = \frac{1 - \tan A \tan B}{\tan A + \tan B}$

$\displaystyle = \frac{1 - \frac{a}{b}}{a}$

$\displaystyle = \frac{b-a}{ab}$

$\displaystyle = \frac{1}{a} - \frac{1}{b}$

$\displaystyle \\$

$\displaystyle \text{Question 24: If } x \text{ lies in the first quadrant and } \cos x = \frac{8}{17} , \text{ then prove that:}$

$\displaystyle \cos \Big( \frac{\pi}{6} + x \Big) + \cos \Big( \frac{\pi}{4} - x \Big) + \cos \Big( \frac{2\pi}{3} - x \Big) = \Big( \frac{\sqrt{3}-1}{2} + \frac{1}{\sqrt{2}} \Big) \frac{23}{17}$

$\displaystyle \text{Given } \cos x = \frac{8}{17}$

$\displaystyle \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \Big( \frac{8}{17} \Big)^2 } = \sqrt{\frac{225}{289}} = \frac{15}{17}$

$\displaystyle \cos \Big( \frac{\pi}{6} + x \Big) + \cos \Big( \frac{\pi}{4} - x \Big) + \cos \Big( \frac{2\pi}{3} - x \Big)$

$\displaystyle = \Big[ \cos \frac{\pi}{6} \cos x - \sin \frac{\pi}{6} \sin x \Big] + \Big[ \cos \frac{\pi}{4} \cos x + \sin \frac{\pi}{4} \sin x \Big] + \Big[ \cos \frac{2\pi}{3} \cos x + \sin \frac{2\pi}{3} \sin x \Big]$

$\displaystyle = \Big[ \cos \frac{\pi}{6} + \cos \frac{\pi}{4} + \cos \frac{2\pi}{3} \Big] \cos x + \Big[ - \sin \frac{\pi}{6} + \sin \frac{\pi}{4} + \sin \frac{2\pi}{3} \Big] \sin x$

$\displaystyle = \Big[ \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} + \cos \Big( \frac{\pi}{2} + \frac{\pi}{6} \Big) \Big] \times \frac{8}{17} + \Big[ - \frac{1}{2} + \frac{1}{\sqrt{2}} + \sin \Big( \frac{\pi}{2} + \frac{\pi}{6} \Big) \Big] \times \frac{15}{17}$

$\displaystyle = \Big[ \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} - \sin \frac{\pi}{6} \Big] \times \frac{8}{17} + \Big[ - \frac{1}{2} + \frac{1}{\sqrt{2}} + \cos \frac{\pi}{6} \Big] \times \frac{15}{17}$

$\displaystyle = \Big[ \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} - \frac{1}{2} \Big] \times \frac{8}{17} + \Big[ - \frac{1}{2} + \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \Big] \times \frac{15}{17}$

$\displaystyle = \Big( \frac{\sqrt{3}-1}{2} + \frac{1}{\sqrt{2}} \Big) ( \frac{8}{17} + \frac{15}{17} )$

$\displaystyle = \Big( \frac{\sqrt{3}-1}{2} + \frac{1}{\sqrt{2}} \Big) \Big( \frac{23}{17} \Big)$

$\displaystyle \\$

$\displaystyle \text{Question 25: If } \tan x + \tan \Big( x + \frac{\pi}{3} \Big) + \tan \Big( x + \frac{2\pi}{3} \Big) = 3 , \\ \\ \text{ then prove that } \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x} = 1$

Given

$\displaystyle \tan x + \tan \Big( x + \frac{\pi}{3} \Big) + \tan \Big( x + \frac{2\pi}{3} \Big) = 3$

$\displaystyle \Rightarrow \tan x + \frac{\tan x + \tan \frac{\pi}{3} }{1 - \tan x \tan \frac{\pi}{3} } + \frac{\tan x + \tan \frac{2\pi}{3} }{1 - \tan x \tan \frac{2\pi}{3} } = 3$

$\displaystyle \Rightarrow \tan x + \frac{\tan x + \sqrt{3} }{1 - \sqrt{3}\tan x } + \frac{\tan x + \tan \Big( \frac{\pi}{2} + \frac{\pi}{3} \Big) }{1 - \tan x \tan \Big( \frac{\pi}{2} + \frac{\pi}{3} \Big) } = 3$

$\displaystyle \Rightarrow \tan x + \frac{\tan x + \sqrt{3} }{1 - \sqrt{3}\tan x } + \frac{\tan x - \cot \frac{\pi}{3} }{1 - \tan x \cot \frac{\pi}{3} } = 3$

$\displaystyle \Rightarrow \tan x + \frac{\tan x + \sqrt{3} }{1 - \sqrt{3}\tan x } + \frac{\tan x - \sqrt{3} }{1 + \sqrt{3}\tan x } = 3$

$\displaystyle \Rightarrow \tan x + \frac{(\tan x + \sqrt{3})(1 + \sqrt{3}\tan x) + (\tan x - \sqrt{3})(1 - \sqrt{3}\tan x)}{(1 - \sqrt{3}\tan x )(1 + \sqrt{3}\tan x)} = 3$

$\displaystyle \Rightarrow \tan x + \frac{\tan x +\sqrt{3} + \sqrt{3} \tan^2 x + 3 \tan x + \tan x - \sqrt{3} - \sqrt{3} \tan^2 x + 3 \tan x }{1 - 3 \tan^2 x} = 3$

$\displaystyle \Rightarrow \tan x + \frac{8 \tan x}{1-3 \tan^2 x} = 3$

$\displaystyle \Rightarrow \frac{\tan x (1-3 \tan^2 x) + 8 \tan x }{1-3 \tan^2 x} = 3$

$\displaystyle \Rightarrow \frac{\tan x - 3 \tan^2 x + 8 \tan x }{1-3 \tan^2 x} = 3$

$\displaystyle \Rightarrow \frac{9 \tan x - 3 \tan^3 x }{1-3 \tan^2 x} = 3$

$\displaystyle \Rightarrow \frac{3 \tan x - \tan^3 x }{1-3 \tan^2 x} = 1$. Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 26: If } \sin (\alpha + \beta) = 1 \text{ and } \sin (\alpha - \beta) = \frac{1}{2}, \\ \\ \text{where } 0 \leq \alpha, \beta \leq \frac{\pi}{2} , \text{ then find the values of } \tan ( \alpha + 2 \beta) \text{ and } \tan ( 2\alpha + \beta)$

Given:

$\displaystyle \sin (\alpha + \beta) = 1 \Rightarrow \alpha + \beta = \frac{\pi}{2}$ … … … i)

$\displaystyle \sin (\alpha - \beta) = \frac{1}{2} \Rightarrow \alpha - \beta = \frac{\pi}{6}$ … … … ii)

Solving i) and ii) we get

$\displaystyle 2 \alpha = \frac{\pi}{2} + \frac{\pi}{6} \Rightarrow \alpha = \frac{\pi}{3}$

Also $\displaystyle \beta = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$

$\displaystyle \tan ( \alpha + 2 \beta) = \tan ( \frac{\pi}{3} + 2 \times \frac{\pi}{6} ) = \tan \frac{\pi}{3} = \tan ( \frac{\pi}{2} + \times \frac{\pi}{6} ) = - \cot \frac{\pi}{6} = -\sqrt{3}$

$\displaystyle \tan ( 2\alpha + \beta) = \tan (2 \frac{\pi}{3} + \frac{\pi}{6} ) = \tan \frac{5\pi}{6} = \tan ( \frac{\pi}{2} + \times \frac{\pi}{3} ) = - \cot \frac{\pi}{3} = \frac{-1}{\sqrt{3}}$

$\displaystyle \\$

$\displaystyle \text{Question 27: If } \alpha \text{ and } \beta$ are two different values of $\displaystyle x$ lying between $\displaystyle 0 \text{ and } 2\pi$ which satisfy the equation $\displaystyle 6 \cos x + 8 \sin x = 9$, find the value of $\displaystyle \sin (\alpha + \beta)$

Given

$\displaystyle 6 \cos x + 8 \sin x = 9$

Eliminating $\displaystyle \sin x$

$\displaystyle \Rightarrow 8 \sin x = 9 - 6 \cos x$

Squaring both sides

$\displaystyle \Rightarrow (8 \sin x)^2 = (9 - 6 \cos x)^2$

$\displaystyle \Rightarrow 64 \sin^2 x = 81 + 36 \cos^2 x - 108 \cos x$

$\displaystyle \Rightarrow 64 (1 - \cos^2 x) = 81 + 36 \cos^2 x - 108 \cos x$

$\displaystyle \Rightarrow 64 - 64 \cos^2 x = 81 + 36 \cos^2 x - 108 \cos x$

$\displaystyle \Rightarrow 100 \cos^2 x - 108 \cos x + 17 = 0$

$\displaystyle \text{If } \alpha \text{ and } \beta$ are roots of equation

$\displaystyle \Rightarrow \cos \alpha \times \cos \beta = \frac{17}{100}$

Similarly

$\displaystyle 6 \cos x + 8 \sin x = 9$

Eliminating $\displaystyle \cos x$

$\displaystyle \Rightarrow 6 \cos x = 9 - 8 \sin x$

Squaring both sides

$\displaystyle \Rightarrow (6 \cos x)^2 = (9 - 8 \sin x)^2$

$\displaystyle \Rightarrow 36 \cos^2 x = 81 + 64 \sin^2 x - 144 \sin x$

$\displaystyle \Rightarrow 36 (1 - \sin^2 x) = 81 + 64 \sin^2 x - 144 \sin x$

$\displaystyle \Rightarrow 36 - 36 \sin^2 x = 81 + 64 \sin^2 x - 144 \sin x$

$\displaystyle \Rightarrow 100 \sin^2 x - 144 \sin x + 17 = 0$

$\displaystyle \text{If } \alpha \text{ and } \beta$ are roots of equation

$\displaystyle \Rightarrow \sin \alpha \times \sin \beta = \frac{17}{100}$

$\displaystyle \text{Now } \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \frac{17}{100} - \frac{45}{100} = \frac{-28}{100} = \frac{-7}{25}$

$\displaystyle \sin (\alpha + \beta) = \sqrt{1 - [\cos(\alpha + \beta) ]^2} = \sqrt{1 - \Big[ \frac{-7}{25} \Big]^2} = \sqrt{1 - \frac{49}{625} } = \sqrt{\frac{576}{625}} = \frac{24}{25}$

$\displaystyle \\$

$\displaystyle \text{Question 28: If } \sin \alpha + \sin \beta = a \text{ and } \cos \alpha + \cos \beta = b$, show that

$\displaystyle \text{ (i) } \sin (\alpha + \beta) = \frac{2ab}{a^2 + b^2}$

$\displaystyle \text{ (ii) } \cos (\alpha + \beta) = \frac{b^2 - a^2}{b^2 + a^2}$

Given, $\displaystyle \sin \alpha + \sin \beta = a \text{ and } \cos \alpha + \cos \beta = b$

$\displaystyle \therefore a^2 + b^2 = (\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2$

$\displaystyle \Rightarrow a^2 + b^2= \sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta + \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta$

$\displaystyle \Rightarrow a^2 + b^2= 2 + 2 (\sin \alpha \sin \beta + \cos \alpha \cos \beta)$

$\displaystyle \Rightarrow a^2 + b^2= 2 + 2 \cos ( \alpha + \beta )$

$\displaystyle \text{Similarly, } \therefore b^2 - a^2 = (\cos \alpha + \cos \beta)^2 - (\sin \alpha + \sin \beta)^2$

$\displaystyle \Rightarrow b^2 - a^2 = \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta - \sin^2 \alpha - \sin^2 \beta - 2 \sin \alpha \sin \beta$

$\displaystyle \Rightarrow b^2 - a^2 = (\cos^2 \alpha - \sin^2 \beta) + (\cos^2 \beta - \sin^2 \alpha) + 2( cos \alpha \cos \beta - \sin \alpha \sin \beta)$

$\displaystyle \Rightarrow b^2 - a^2 = \cos (\alpha + \beta) \cos(\alpha - \beta) + \cos ( \beta + \alpha) \cos ( \beta - \alpha) + 2 \cos (\alpha + \beta)$

$\displaystyle \because \cos (\beta - \alpha) = \cos [ - ( \alpha - \beta ) ] = \cos ( \alpha - \beta )$

$\displaystyle \Rightarrow b^2 - a^2 = \cos (\alpha + \beta) [ 2 \cos ( \alpha - \beta ) + 2]$

$\displaystyle \Rightarrow b^2 - a^2 = \cos (\alpha + \beta) [ a^2 + b^2 ]$

$\displaystyle \Rightarrow \cos ( \alpha + \beta ) = \frac{b^2 - a^2}{b^2 + a^2}$

$\displaystyle \sin ( \alpha + \beta ) = \sqrt{1 - [\cos ( \alpha + \beta )]^2 }$

$\displaystyle = \sqrt{1 - \Big( \frac{b^2 - a^2}{b^2 + a^2} \Big)^2 }$

$\displaystyle = \sqrt{ \frac{b^4 + a^4 + 2 b^2 a^2 - b^4 - a^4 + 2 b^2a^2}{(b^2 + a^2)^2} }$

$\displaystyle = \frac{2ba}{b^2 + a^2}$

$\displaystyle \\$

Question 29: Prove that:

$\displaystyle \text{ (i) } \frac{1}{\sin (x-a) \sin (x-b) } = \frac{\cot (x-a) - \cot (x-b)}{\sin (a-b)}$

$\displaystyle \text{ (ii) } \frac{1}{\sin (x-a) \cos (x-b) } = \frac{\cot (x-a) + \tan (x-b)}{\cos (a-b)}$

$\displaystyle \text{ (iii) } \frac{1}{\cos (x-a) \cos (x-b) } = \frac{\tan (x-b) - \tan (x-a)}{\sin (a-b)}$

$\displaystyle \text{(i) } \text{LHS } = \frac{1}{\sin (x-a) \sin (x-b) }$

$\displaystyle = \frac{1}{\sin(a-b)} \Big[ \frac{\sin(a-b)}{\sin (x-a) \sin (x-b)} \Big]$

$\displaystyle = \frac{1}{\sin(a-b)} \Big[ \frac{\sin [(x-b) - (x-a) ] }{\sin (x-a) \sin (x-b)} \Big]$

$\displaystyle = \frac{1}{\sin(a-b)} \Big[ \frac{\sin (x-b) \cos (x-a) - \cos (x-b) \sin (x-a) }{\sin (x-a) \sin (x-b)} \Big]$

$\displaystyle = \frac{1}{\sin(a-b)} \Big[ \frac{\sin (x-b) \cos (x-a)}{\sin (x-a) \sin (x-b)} - \frac{\cos (x-b) \sin (x-a)}{\sin (x-a) \sin (x-b)} \Big]$

$\displaystyle = \frac{1}{\sin(a-b)} [ \cot (x-a) - \cot ( x- b) ]$

$\displaystyle = \frac{ \cot (x-a) - \cot ( x- b)}{\sin(a-b)}$

$\displaystyle =$ RHS. Hence proved.

$\displaystyle \text{(ii) } \text{LHS } = \frac{1}{\sin (x-a) \cos (x-b) }$

$\displaystyle = \frac{1}{\cos(a-b)} \Big[ \frac{\cos(a-b)}{\sin (x-a) \cos (x-b)} \Big]$

$\displaystyle = \frac{1}{\cos(a-b)} \Big[ \frac{\cos [(x-b) - (x-a) ] }{\sin (x-a) \cos (x-b)} \Big]$

$\displaystyle = \frac{1}{\cos(a-b)} \Big[ \frac{\cos (x-b) \cos (x-a) + \sin (x-b) \sin (x-a) }{\sin (x-a) \cos (x-b)} \Big]$

$\displaystyle = \frac{1}{\cos(a-b)} \Big[ \frac{\cos (x-b) \cos (x-a)}{\sin (x-a) \cos (x-b)} + \frac{\sin (x-b) \sin (x-a)}{\sin (x-a) \cos (x-b)} \Big]$

$\displaystyle = \frac{1}{\cos(a-b)} [ \cot (x-a) + \tan ( x- b) ]$

$\displaystyle = \frac{ \cot (x-a) + \tan ( x- b)}{\cos(a-b)}$

$\displaystyle =$ RHS. Hence proved.

$\displaystyle \text{(iii) } \text{LHS } = \frac{1}{\cos (x-a) \cos (x-b) }$

$\displaystyle = \frac{1}{\sin(a-b)} \Big[ \frac{\sin(a-b)}{\cos (x-a) \cos (x-b)} \Big]$

$\displaystyle = \frac{1}{\sin(a-b)} \Big[ \frac{\sin [(x-b) - (x-a) ] }{\cos (x-a) \cos (x-b)} \Big]$

$\displaystyle = \frac{1}{\sin(a-b)} \Big[ \frac{\sin (x-b) \cos (x-a) - \cos (x-b) \sin (x-a) }{\cos (x-a) \cos (x-b)} \Big]$

$\displaystyle = \frac{1}{\sin(a-b)} \Big[ \frac{\sin (x-b) \cos (x-a)}{\cos (x-a) \cos (x-b)} - \frac{\cos (x-b) \sin (x-a)}{\cos (x-a) \cos (x-b)} \Big]$

$\displaystyle = \frac{1}{\sin(a-b)} [ \tan (x-b) - \tan ( x- a) ]$

$\displaystyle = \frac{ \tan (x-b) - \tan ( x- a)}{\sin(a-b)}$

$\displaystyle =$ RHS. Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 30: If } \sin \alpha \sin \beta - \cos \alpha \cos \beta + 1 = 0$, prove that $\displaystyle 1 + \cot \alpha \tan \beta = 0$

$\displaystyle \text{Given } \sin \alpha \sin \beta - \cos \alpha \cos \beta + 1 = 0$

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

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

$\displaystyle \therefore \sin ( \alpha + \beta ) = \sqrt{ 1 - \cos^2 ( \alpha + \beta ) } = \sqrt{ 1 - 1 } = 0$

$\displaystyle 1 + \cot \alpha \tan \beta = 1 + \frac{\cos \alpha}{\sin \alpha} \times \frac{\sin \beta}{\cos \beta}$

$\displaystyle \Rightarrow 1 + \cot \alpha \tan \beta = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\sin \alpha \cos \beta }$

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

$\displaystyle \Rightarrow 1 + \cot \alpha \tan \beta = 0$ Hence Proved.

$\displaystyle \\$

$\displaystyle \text{Question 31: If } \tan \alpha = x + 1 \text{ and } \tan \beta = x - 1$, show that $\displaystyle 2 \cot (\alpha - \beta) = x^2$

$\displaystyle \text{Given } \tan \alpha = x + 1 \text{ and } \tan \beta = x - 1$

$\displaystyle 2 \cot (\alpha - \beta) = \frac{2}{\tan (\alpha - \beta)}$

$\displaystyle \Rightarrow 2 \cot (\alpha - \beta) = \frac{2 ( 1 + \tan \alpha \tan \beta}{\tan \alpha - \tan \beta}$

$\displaystyle \Rightarrow 2 \cot (\alpha - \beta) = 2 \Big[ \frac{1 + ( x + 1 ) ( x - 1 ) }{(x+ 1 ) - ( x - 1)} \Big]$

$\displaystyle \Rightarrow 2 \cot (\alpha - \beta) = 2 \Big[ \frac{1 +x^2 - 1}{2} \Big]$

$\displaystyle \Rightarrow 2 \cot (\alpha - \beta) = x^2$. Hence proved.

$\displaystyle \\$

Question 32: If angle $\displaystyle \theta$ is divided into two parts such that the tangents of one parts is $\displaystyle \lambda$ times the tangent of other, and $\displaystyle \phi$ is their difference, then show that $\displaystyle \sin \theta = \frac{\lambda - 1}{\lambda + 1} \sin \phi$

Let $\displaystyle \alpha \text{ and } \beta$ be the two parts

$\displaystyle \theta = \alpha + \beta$

$\displaystyle \phi = \alpha - \beta$

$\displaystyle \tan \alpha = \lambda \tan \beta$

$\displaystyle \Rightarrow \frac{\tan \alpha}{\tan \beta} = \frac{\lambda}{1}$

Applying componendo and dividendo

$\displaystyle \Rightarrow \frac{\tan \alpha + \tan \beta}{\tan \alpha - \tan \beta} = \frac{\lambda+ 1}{\lambda - 1}$

$\displaystyle \Rightarrow \frac{ \frac{\sin \alpha}{\cos \alpha} + \frac{\sin \beta}{\cos \beta}}{\frac{\sin \alpha}{\cos \alpha} - \frac{\sin \beta}{\cos \beta}} = \frac{\lambda+ 1}{\lambda - 1}$

$\displaystyle \Rightarrow \frac{\sin \alpha \cos \beta + \sin \beta \cos \alpha}{\sin \alpha \cos \beta - \sin \beta \cos \alpha} = \frac{\lambda+ 1}{\lambda - 1}$

$\displaystyle \Rightarrow \frac{\sin ( \alpha + \beta) }{\sin ( \alpha - \beta) } = \frac{\lambda+ 1}{\lambda - 1}$

$\displaystyle \Rightarrow \frac{\sin \theta }{\sin \phi} = \frac{\lambda+ 1}{\lambda - 1}$

$\displaystyle \Rightarrow \sin \theta = \Big( \frac{\lambda+ 1}{\lambda - 1} \Big) \sin \phi \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 33: If } \tan x = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha} , \text{ then show that } \sin \alpha + \cos \alpha = \sqrt{2} \cos x$

$\displaystyle \text{Given } \tan x = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}$

Dividing both numerator and denominator by $\displaystyle \cos \alpha$

$\displaystyle \Rightarrow \tan \theta = \frac{\tan \alpha - 1}{\tan \alpha + 1}$

$\displaystyle \Rightarrow \tan \theta = \frac{\tan \alpha - \tan \frac{\pi}{4}}{1 + \tan \alpha \tan \frac{\pi}{4}}$

$\displaystyle \Rightarrow \tan \theta = \tan (\alpha - \frac{\pi}{4} )$

$\displaystyle \Rightarrow \theta = \alpha - \frac{\pi}{4}$

$\displaystyle \Rightarrow \cos \theta = \cos (\alpha - \frac{\pi}{4} )$

$\displaystyle \Rightarrow \cos \theta = \cos \alpha \cos \frac{\pi}{4} + \sin \alpha \sin \frac{\pi}{4}$

$\displaystyle \Rightarrow \cos \theta = \frac{\cos \alpha}{\sqrt{2}} + \frac{\sin \alpha}{\sqrt{2}}$

$\displaystyle \Rightarrow \sqrt{2} \cos \theta = \cos \alpha + \sin \alpha$ Hence proved.

$\displaystyle \\$

Question 34: $\displaystyle \text{If } \alpha \text{ and } \beta$ are two solutions of the equation $\displaystyle a \tan x + b \sec x = c$, then find the values of $\displaystyle \sin (\alpha+\beta) \text{ and } \cos (\alpha+\beta)$

$\displaystyle \text{Given } a \tan x + b \sec x = c$

$\displaystyle a \frac{\sin x}{\cos x} + \frac{1}{\cos x} = c$

$\displaystyle a \sin x + b = c \cos x$

Squaring both sides

$\displaystyle (a \sin x + b)^2 = (c \cos x)^2$

$\displaystyle a^2 \sin^2 x+ b^2 + 2 ab \sin x = c^2 \cos^2 x$

$\displaystyle a^2 \sin^2 x+ b^2 + 2 ab \sin x = c^2 (1 - \sin^2 x)$

$\displaystyle (a^2 + c^2) \sin^2 x + 2 ab \sin x + (b^2 - c^2) = 0$

$\displaystyle \text{If } \alpha \text{ and } \beta$ are the roots then

$\displaystyle \therefore \sin \alpha \sin \beta = \frac{b^2 - c^2}{a^2 + c^2}$

$\displaystyle \text{Similarly } a \sin x + b = c \cos x$

$\displaystyle a \sin x = c \cos x - b$

Squaring both sides

$\displaystyle (a \sin x)^2 = (c \cos x - b)^2$

$\displaystyle a^2 \sin^2 x = c^2 \cos^2 x + b^2 - 2bc \cos x$

$\displaystyle (c^2 + a^2) \cos^2 x - 2bc \cos x + (b^2 - a^2) = 0$

$\displaystyle \text{If } \alpha \text{ and } \beta$ are the roots then

$\displaystyle \therefore \cos \alpha \cos \beta = \frac{b^2 - a^2}{c^2 + a^2}$

Now $\displaystyle \cos( \alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

$\displaystyle = \frac{b^2 - a^2}{c^2 + a^2} - \frac{b^2 - c^2}{a^2 + c^2}$

$\displaystyle = \frac{b^2 - a^2 - b^2 + c^2 }{a^2 + c^2}$

$\displaystyle = \frac{ c^2 - a^2 }{a^2 + c^2}$

$\displaystyle \sin ( \alpha + \beta) = \sqrt{1 - \cos^2 ( \alpha + \beta) }$

$\displaystyle = \sqrt{1 - \Big( \frac{ c^2 - a^2 }{a^2 + c^2} \Big) }$

$\displaystyle = \sqrt{ \frac{c^4 + a^4 + 2 c^2 a^2- c^4 - a^4 + 2 c^2 a^2}{(c^2 + a^2)^2} }$

$\displaystyle = \frac{2ac}{c^2 + a^2}$