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

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

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

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

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

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

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

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

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

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Question 2:

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

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

a) Given $\sin A =$ $\frac{12}{13}$ and $\sin B =$ $\frac{4}{5}$, 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Question 6: If $\sin A =$ $\frac{1}{2}$ and $\cos B =$ $\frac{\sqrt{3}}{2}$, where $\frac{\pi}{2}$ $< A < \pi$  and $0 < B <$ $\frac{\pi}{2}$, find the following:   i) $\tan (A+B)$   ii) $\tan (A-B)$

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

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

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

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

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

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

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

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

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Question 7: Evaluate the following:

i) $\sin 78^o \cos 18^o - \cos 78^o \sin 18^o$   ii) $\cos 47^o \cos 13^o - \sin 47^o \sin 13^o$

iii) $\sin 36^o \cos 9^o + \cos 36^o \sin 9^o$   iv)$\cos 80^o \cos 20^o + \sin 80^o \sin 20^o$

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

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

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

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

$\\$

Question 8: If $\cos A = -$ $\frac{12}{13}$ 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:

i) $\sin (A+B)$   ii) $\cos (A+B)$   iii) $\tan (A + B)$

Given $\cos A = -$ $\frac{12}{13}$ and $\cot B =$ $\frac{24}{7}$, where A lies in the second quadrant and B in the third quadrant

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

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

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

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

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

$\\$

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

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

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

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

$=$ RHS. Hence proved.

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

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

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

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

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

$=$ RHS. Hence proved.

$\\$

Question 11:

i) $\frac{\cos 11^o + \sin 11^o}{\cos 11^o - \sin 11^o}$ $= \tan 56^o$

ii) $\frac{\cos 9^o + \sin 9^o}{\cos 9^o - \sin 9^o}$ $= \tan 54^o$

iii) $\frac{\cos 8^o - \sin 8^o}{\cos 8^o + \sin 8^o}$ $= \tan 37^o$

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

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

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

$\\$

Question 12:

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

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

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

i) 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: $\sin (A + B) = \sin A \cos B + \cos A + \sin B$

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

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

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

$= 1 =$ RHS. Hence proved.

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

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

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

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

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

iii) 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: $\sin (A + B) = \sin A \cos B + \cos A + \sin B$

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

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

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

$= 1 =$ RHS. Hence proved.

$\\$

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

LHS $=$ $\frac{\tan 69^o + \tan 66^o}{1 - \tan 69^o \tan 66^o}$

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

$= \tan (69^o+66^o) = \tan 135^o - \cot 45^o = -1 =$ RHS. Hence proved.

$\\$

Question 14: i) If $\tan A =$ $\frac{5}{6}$ and $\tan B =$ $\frac{1}{11}$, prove that $A + B =$ $\frac{\pi}{4}$

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

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

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

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

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

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

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

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

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

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

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

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

$\\$

Question 15: Prove that

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

ii) $\sin^2 (n+1) A - \sin^2 nA = \sin(2n+1) A \sin A$

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

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

Lets calculate $\sin^2$ $\frac{\pi}{12} = \sin^2 15^o$

$\sin 15^o = \sin ( 45^o- 30^o)$

$= \sin 45^o \cos 30^o - \cos 45^o \sin 30^o$

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

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

Substituting it back

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

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

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

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

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

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

$=$ RHS. Hence proved.

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

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

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

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

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

$=$ RHS. Hence Proved.

$\\$

Question 16: Prove that:

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

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

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

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

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

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

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

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

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

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

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

$= 0 =$ RHS. Hence proved.

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

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

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

$= 0 =$ RHS. Hence proved.

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

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

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

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

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

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

$= \sin^2 B =$ LHS. Hence Proved

v) LHS $= \cos^2 A + \cos^2 B - 2 \cos A \cos B \cos (A+ B)$

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

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

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

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

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

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

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

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

$= - \cos^2 ( A + B) + 1 = \sin^2 ( A + B) =$ RHS. Hence Proved.

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

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

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

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

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

$\\$

Question 17: Prove that:

i) $\tan 8x - \tan 6x - \tan 2x = \tan 8x \tan 6x \tan 2x$

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

iii) $\tan 36^o + \tan 9^o + \tan 36^o \tan 9^o = 1$

iv) $\tan 13x - \tan 9x - \tan 4x = \tan 13x \tan 9x \tan 4x$

i) $8x = 6x + 2x$

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

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

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

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

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

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

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

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

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

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

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

iii)  $45^o = 36^o + 9^o$

$\Rightarrow \tan 45^o = \tan (36^o + 9^o)$

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

$\Rightarrow 1 ( 1- \tan 36^o \tan 9^o ) = \tan 36^o + \tan 9^o$

$\Rightarrow 1 - \tan 36^o \tan 9^o = \tan 36^o + \tan 9^o$

$\Rightarrow 1 - \tan 36^o - \tan 9^o = \tan 36^o \tan 9^o$

iv)  $13x = 9x + 4x$

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

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

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

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

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

$\\$

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

RHS $= \tan 3x \tan x$

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

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

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

$=$ LHS. Hence proved.

$\\$

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

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

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

Using Componendo and Dividendo

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

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

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

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

$\\$

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

$\tan A = x \tan B$

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

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

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

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

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

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

Hence proved.

$\\$

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

Given $\tan (A+B) = x$ and $\tan (A-B) = y$

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

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

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

Similarly,

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

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

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

$\\$

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

Given $\cos A + \sin B = m$ and $\sin A + \cos B = n$

$m^2 + n^2 - 2$

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

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

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

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

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

$= 2 \sin ( A + B)$   Hence proved.

$\\$

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

Given $\tan A + \tan B = a$ and $\cot A + \cot B = b$

$\cot A + \cot B = b$

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

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

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

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

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

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

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

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

$\\$

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

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

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

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

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

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

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

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

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

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

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

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

$\\$

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

Given

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

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

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

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

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

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

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

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

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

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

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

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

$\\$

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

Given:

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

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

Solving i) and ii) we get

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

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

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

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

$\\$

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

Given

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

Eliminating $\sin x$

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

Squaring both sides

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

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

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

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

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

If $\alpha$ and $\beta$ are roots of equation

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

Similarly

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

Eliminating $\cos x$

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

Squaring both sides

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

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

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

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

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

If $\alpha$ and $\beta$ are roots of equation

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

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

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

$\\$

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

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

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

Given, $\sin \alpha + \sin \beta = a$ and $\cos \alpha + \cos \beta = b$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\\$

Question 29: Prove that:

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

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

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

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

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

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

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

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

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

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

$=$ RHS. Hence proved.

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

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

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

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

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

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

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

$=$ RHS. Hence proved.

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

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

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

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

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

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

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

$=$ RHS. Hence proved.

$\\$

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

Given $\sin \alpha \sin \beta - \cos \alpha \cos \beta + 1 = 0$

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

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

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

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

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

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

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

$\\$

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

Given $\tan \alpha = x + 1$ and $\tan \beta = x - 1$

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

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

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

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

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

$\\$

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

Let $\alpha$ and $\beta$ be the two parts

$\theta = \alpha + \beta$

$\phi = \alpha - \beta$

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

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

Applying componendo and dividendo

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Question 34: If