Class 11: Transformation Formulae – Exercise 8.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Express each of the following as the product of sines and cosines:

i) $\sin 12 x + \sin 4x$ ii) $\sin 5x - \sin x$ iii) $\cos 12 x + \cos 8x$

iv) $\cos 12 x - \cos 4x$ v) $\sin 2x + \cos 4x$

Answer:

i) $\sin 12 x + \sin 4x = 2 \sin \Big($ $\frac{12x + 4x}{2}$ $\Big) \cos \Big($ $\frac{12x - 4x}{2}$ $\Big) = 2 \sin 8x \cos 4x$

ii) $\sin 5x - \sin x = 2 \cos \Big($ $\frac{5x + x}{2}$ $\Big) \sin \Big($ $\frac{5x - x}{2}$ $\Big) = 2 \cos 3x \sin 2x$

iii) $\cos 12 x + \cos 8x = 2 \cos \Big($ $\frac{12x + 8x}{2}$ $\Big) \cos \Big($ $\frac{12x - 8x}{2}$ $\Big) = 2 \cos 10x \cos 2x$

iv) $\cos 12 x - \cos 4x = -2 \sin \Big($ $\frac{12x + 4x}{2}$ $\Big) \sin \Big($ $\frac{12x - 4x}{2}$ $\Big) = -2 \sin 8x \sin 4x$

v) $\sin 2x + \cos 4x = \sin 2x + \sin (90^o - 4x)$

$= 2 \sin \Big($ $\frac{2x + 90^o - 4x}{2}$ $\Big) \cos \Big($ $\frac{2x - 90^o + 4x}{2}$ $\Big)$

$= 2 \sin (45^o - x) \cos ( 3x - 45^o)$

$= 2 \sin (45^o - x) \cos ( 45^o- 3x)$

$\\$

Question 2: Prove that:

i) $\sin 38^o + \sin 22^o = \sin 82^o$ ii) $\cos 100^o + \cos 20^o = \cos 40^o$

iii) $\sin 50^o + \sin 10^o = \cos 20^o$ iv) $\sin 23^o + \sin 37^o = \cos 7^o$

v) $\sin 105^o + \cos 105^o = \cos 45^o$ vi) $\sin 40^o + \sin 20^o = \cos 10^o$

Answer:

i) LHS $=\sin 38^o + \sin 22^o$

$=2 \sin \Big($ $\frac{38^o+22^o}{2}$ $\Big) \cos \Big($ $\frac{38^o-22^o}{2}$ $\Big)$

$= 2 \sin 30^o \cos 8^o$

$= 2 \times$ $\frac{1}{2}$ $\cos 8^o$

$= \cos (90^o-82^o) = \sin 82^o =$ RHS. Hence proved.

ii) LHS $=\cos 100^o + \cos 20^o$

$=2 \cos \Big($ $\frac{100^o+20^o}{2}$ $\Big) \cos \Big($ $\frac{100^o-20^o}{2}$ $\Big)$

$= 2 \cos 60^o \cos 40^o$

$= 2 \times$ $\frac{1}{2}$ $\cos 40^o$

$= \cos 40^o =$ RHS. Hence proved.

iii) LHS $=\sin 50^o + \sin 10^o$

$=2 \sin \Big($ $\frac{50^o+10^o}{2}$ $\Big) \cos \Big($ $\frac{50^o-10^o}{2}$ $\Big)$

$= 2 \sin 30^o \cos 20^o$

$= 2 \times$ $\frac{1}{2}$ $\cos 20^o$

$= \cos 20^o =$ RHS. Hence proved.

iv) LHS $=\sin 23^o + \sin 37^o$

$=2 \sin \Big($ $\frac{23^o+37^o}{2}$ $\Big) \cos \Big($ $\frac{23^o-37^o}{2}$ $\Big)$

$= 2 \sin 30^o \cos (-7^o)$

$= 2 \times$ $\frac{1}{2}$ $\cos 7^o$

$= \cos 7^o =$ RHS. Hence proved.

v) LHS $=\sin 105^o + \cos 105^o$

$=\sin 105^o + \cos (90^o+ 15^o)$

$=\sin 105^o - \sin 15^o$

$=2 \sin \Big($ $\frac{105^o-15^o}{2}$ $\Big) \cos \Big($ $\frac{105^o+15^o}{2}$ $\Big)$

$= 2 \sin 45^o \cos 60^o$

$= 2 \times$ $\frac{1}{\sqrt{2}}$ $\times$ $\frac{1}{2}$

$=$ $\frac{1}{\sqrt{2}}$ $= \cos 45^o =$ RHS. Hence proved.

vi) LHS $=\sin 40^o + \sin 20^o$

$=2 \sin \Big($ $\frac{40^o+20^o}{2}$ $\Big) \cos \Big($ $\frac{20^o-40^o}{2}$ $\Big)$

$= 2 \sin 30^o \cos (-10^o)$

$= 2 \times$ $\frac{1}{2}$ $\cos 10^o$

$= \cos 10^o =$ RHS. Hence proved.

$\\$

Question 3: Prove that:

i) $\cos 55^o + \cos 65^o + \cos 175^o = 0$ ii) $\sin 50^o - \sin 70^o + \sin 10^o = 0$

iii) $\cos 80^o + \cos 40^o - \cos 20^o = 0$ v) $\cos 20^o + \cos 100^o + \cos 140^o = 0$

v) $\sin$ $\frac{5\pi}{18}$ $- \cos$ $\frac{4\pi}{9}$ $= \sqrt{3} \sin$ $\frac{\pi}{9}$ vi) $\cos$ $\frac{\pi}{12}$ $- \sin$ $\frac{\pi}{12}$ $=$ $\frac{1}{\sqrt{2}}$

vii) $\sin 80^o - \cos 70^o = \cos 50^o$ viii) $\sin 51^o + \cos 81^o = \cos 21^o$

Answer:

i) LHS $= \cos 55^o + \cos 65^o + \cos 175^o$

$= \cos 55^o + \cos 65^o + \cos ( 180^o - 5^o)$

$= \cos 55^o + \cos 65^o - \cos 5^o$

$= 2 \cos$ $\frac{55^o+65^o}{2}$ $\cos$ $\frac{55^o-65^o}{2}$ $- \cos 5^o$

$=2 \cos 60^o \cos 5^o - \cos 5^o$

$= 2 \times$ $\frac{1}{2}$ $\cos 5^o - \cos 5^o$

$= \cos 5^o - \cos 5^o = 0 =$ RHS. Hence proved.

ii) LHS $= \sin 50^o - \sin 70^o + \sin 10^o$

$= 2 \sin$ $\frac{50^o-70^o}{2}$ $\cos$ $\frac{50^o+70^o}{2}$ $+ \sin 10^o$

$=2 \sin (-10^o) \cos 60^o + \sin 10^o$

$= -2 \times$ $\frac{1}{2}$ $\sin 10^o + \sin 10^o$

$= -\sin 10^o + \sin 10^o = 0 =$ RHS. Hence proved.

iii) LHS $= \cos 80^o + \cos 40^o - \cos 20^o$

$= 2 \cos$ $\frac{80^o+40^o}{2}$ $\cos$ $\frac{80^o-40^o}{2}$ $- \cos 20^o$

$=2 \cos 60^o \cos 20^o - \cos 20^o$

$= 2 \times$ $\frac{1}{2}$ $\cos 20^o - \cos 20^o$

$= \cos 20^o - \cos 20^o = 0 =$ RHS. Hence proved.

iv) LHS $= \cos 20^o + \cos 100^o + \cos 140^o$

$= 2 \cos$ $\frac{20^o+100^o}{2}$ $\cos$ $\frac{20^o-100^o}{2}$ $+ \cos (180^o - 40^o)$

$=2 \cos 60^o \cos 40^o - \cos 40^o$

$= 2 \times$ $\frac{1}{2}$ $\cos 40^o - \cos 40^o$

$= \cos 40^o - \cos 40^o = 0 =$ RHS. Hence proved.

v) LHS $= \sin$ $\frac{5\pi}{18}$ $- \cos$ $\frac{4\pi}{9}$

$= \sin 50^o - \cos 80^o = \sin 50^o - \cos ( 90^o-10^o) = \sin 50^o - \sin 10^o$

$= 2 \sin$ $\frac{50^o-10^o}{2}$ $\cos$ $\frac{50^o+10^o}{2}$

$=2 \sin 20^o \cos 30^o$

$= 2 \times$ $\frac{\sqrt{3}}{2}$ $\sin 20^o$

$= \sqrt{3} \sin 20^o = \sqrt{3} \sin$ $\frac{\pi}{9}$ $=$ RHS. Hence proved.

vi) LHS $= \cos$ $\frac{\pi}{12}$ $- \sin$ $\frac{\pi}{12}$

$= \cos 15^o - \sin 15^o = \cos 15^o - \sin ( 90^o-75^o) = \cos 15^o - \cos 75^o$

$= 2 \sin$ $\frac{15^o+75^o}{2}$ $\sin$ $\frac{75^o-15^o}{2}$

$= 2 \sin 45^o \sin 30^o$

$= 2 \times$ $\frac{1}{\sqrt{2}}$ $\sin 30^o$

$= 2 \times$ $\frac{1}{\sqrt{2}}$ $\times$ $\frac{1}{2}$ $=$ $\frac{1}{\sqrt{2}}$ $=$ RHS. Hence proved.

vii) LHS $= \sin 80^o - \cos 70^o = \sin 80^o - \cos ( 90^o-20^o) = \sin 80^o - \sin 20^o$

$= 2 \sin$ $\frac{80^o-20^o}{2}$ $\cos$ $\frac{80^o+20^o}{2}$

$=2 \sin 30^o \cos 50^o$

$= 2 \times$ $\frac{1}{2}$ $\cos 50^o$

$= \cos 50^o =$ RHS. Hence proved.

viii) LHS $= \sin 51^o + \cos 81^o = \sin 51^o + \cos ( 90^o-9^o) = \sin 51^o + \sin 9^o$

$= 2 \sin$ $\frac{51^o+9^o}{2}$ $\cos$ $\frac{51^o-9^o}{2}$

$=2 \sin 30^o \cos 21^o$

$= 2 \times$ $\frac{1}{2}$ $\cos 21^o$

$= \cos 21^o =$ RHS. Hence proved.

$\\$

Question 4: Prove that:

i) $\cos \Big($ $\frac{3\pi}{4}$ $+x \Big) - \cos \Big($ $\frac{3\pi}{4}$ $-x \Big) = -\sqrt{2} \sin x$

ii) $\cos \Big($ $\frac{\pi}{4}$ $+x \Big) + \cos \Big($ $\frac{\pi}{4}$ $-x \Big) = -\sqrt{2} \cos x$

Answer:

i) LHS $= \cos \Big($ $\frac{3\pi}{4}$ $+x \Big) - \cos \Big($ $\frac{3\pi}{4}$ $-x \Big)$

$= - \Big[ \cos \Big($ $\frac{3\pi}{4}$ $-x \Big) - \cos \Big($ $\frac{3\pi}{4}$ $+x \Big) \Big]$

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

$= - \Big[ - 2 \sin$ $\frac{3\pi}{4}$ $\sin (-x) \Big]$

$= - 2 \sin \Big($ $\frac{\pi}{2}$ $+$ $\frac{\pi}{4}$ $\Big) \sin x$

$= -2 \cos$ $\frac{\pi}{4}$ $\sin x = - \sqrt{2} \sin x =$ RHS. Hence proved.

ii) LHS $= \cos \Big($ $\frac{\pi}{4}$ $+x \Big) + \cos \Big($ $\frac{\pi}{4}$ $-x \Big)$

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

$= 2 \cos$ $\frac{\pi}{4}$ $\cos x$

$= \sqrt{2} \cos x =$ RHS. Hence proved.

$\\$

Question 5: Prove that:

i) $\sin 65^o + \cos 65^o = \sqrt{2} \cos 20^o$ ii) $\sin 47^o +\cos 77^o = \cos 17^o$

Answer:

i) LHS $= \sin 65^o + \cos 65^o$

$= \sin 65^o + \cos ( 90^o-25^o)$

$= \sin 65^o + \sin 25^o$

$= 2 \sin \Big($ $\frac{65^o+ 25^o}{2}$ $\Big) \cos \Big($ $\frac{65^o- 25^o}{2}$ $\Big)$

$= 2 \sin 45^o \cos 20^o$

$= 2 \times$ $\frac{1}{\sqrt{2}}$ $\cos 20^o$

$= \sqrt{2} \cos 20^o =$ RHS. Hence proved.

ii) LHS $= \sin 47^o +\cos 77^o$

LHS $= \sin 47^o + \cos 77^o$

$= \sin 47^o + \cos ( 90^o-13^o)$

$= \sin 47^o + \sin 13^o$

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

$= 2 \sin 30^o \cos 17^o$

$= 2 \times$ $\frac{1}{2}$ $\cos 17^o$

$= \cos 17^o =$ RHS. Hence proved.

$\\$

Question 6: Prove that:

i) $\cos 3A + \cos 5A + \cos 7A + \cos 15A = 4 \cos 4A \cos 5A \cos 6A$

ii) $\cos A + \cos 3A + \cos 5A + \cos 7A = 4 \cos A \cos 2A \cos 4A$

iii) $\sin A + \sin 2A + \sin 4A + \sin 5A = 4 \cos$ $\frac{A}{2}$ $\cos$ $\frac{3A}{2}$ $\sin 3A$

iv) $\sin 3A + \sin 2A - \sin A = 4 \sin A \cos$ $\frac{A}{2}$ $\cos$ $\frac{3A}{2}$

v) $\cos 20^o \cos 100^o + \cos 100^o \cos 140^o - \cos 140^o \cos 200^o = -$ $\frac{3}{4}$

vi) $\sin$ $\frac{x}{2}$ $\sin$ $\frac{7x}{2}$ $+ \sin$ $\frac{3x}{2}$ $\sin$ $\frac{11x}{2}$ $= \sin 2x \sin 5x$

vii) $\cos x \cos$ $\frac{x}{2}$ $- \cos 3x \cos$ $\frac{9x}{2}$ $= \sin 4x \sin$ $\frac{7x}{2}$

Answer:

i) LHS $= \cos 3A + \cos 5A + \cos 7A + \cos 15A$

$= [ \cos 5A + \cos 3A] + [ \cos 15A + \cos 7A ]$

$= 2 \cos \Big($ $\frac{5A + 3A}{2}$ $\Big) \cos \Big($ $\frac{5A - 3A}{2}$ $\Big) + 2 \cos \Big($ $\frac{15A + 7A}{2}$ $\Big) \cos \Big($ $\frac{15A - 7A}{2}$ $\Big)$

$= 2 \cos 4A \cos A + 2 \cos 11 A \cos 4A$

$= 2 \cos 4A [ \cos 11A + \cos A ]$

$= 2 \cos 4A \Big[ 2 \cos \Big($ $\frac{11A + A}{2}$ $\Big) \cos \Big($ $\frac{11A - A}{2}$ $\Big) \Big]$

$= 2 \cos 4A [ 2 \cos 6A \cos 5A ]$

$= 4 \cos 4A \cos 5A \cos 6A =$ RHS. Hence proved.

ii) LHS $= \cos A + \cos 3A + \cos 5A + \cos 7A$

$= [ \cos 3A + \cos A] + [ \cos 7A + \cos 5A ]$

$= 2 \cos \Big($ $\frac{3A + A}{2}$ $\Big) \cos \Big($ $\frac{3A - A}{2}$ $\Big) + 2 \cos \Big($ $\frac{7A + 5A}{2}$ $\Big) \cos \Big($ $\frac{7A - 5A}{2}$ $\Big)$

$= 2 \cos 2A \cos A + 2 \cos 6 A \cos A$

$= 2 \cos A [ \cos 6A + \cos 2A ]$

$= 2 \cos A \Big[ 2 \cos \Big($ $\frac{6A + 2A}{2}$ $\Big) \cos \Big($ $\frac{6A - 2A}{2}$ $\Big) \Big]$

$= 2 \cos A [ 2 \cos 4A \cos 2A ]$

$= 4 \cos A \cos 2A \cos 4A =$ RHS. Hence proved.

iii) LHS $= \sin A + \sin 2A + \sin 4A + \sin 5A$

$= [ \sin 2A + \sin A] + [ \sin 5A + \sin 4A ]$

$= 2 \sin \Big($ $\frac{2A + A}{2}$ $\Big) \cos \Big($ $\frac{2A - A}{2}$ $\Big) + 2 \sin \Big($ $\frac{5A + 4A}{2}$ $\Big) \cos \Big($ $\frac{5A - 4A}{2}$ $\Big)$

$= 2 \sin$ $\frac{3A}{2}$ $\cos$ $\frac{A}{2}$ $+ 2 \sin$ $\frac{9A}{2}$ $\cos$ $\frac{A}{2}$

$= 2 \cos$ $\frac{A}{2}$ $\Big[ \sin$ $\frac{3A}{2}$ $+ \sin$ $\frac{9A}{2}$ $\Big]$

$= 2 \cos$ $\frac{A}{2}$ $\Big[ 2 \sin \Big($ $\frac{\frac{9A}{2} + \frac{3A}{2}}{2}$ $\Big) \cos \Big($ $\frac{\frac{9A}{2} - \frac{3A}{2}}{2}$ $\Big) \Big]$

$= 2 \cos$ $\frac{A}{2}$ $[ 2 \sin 3A \sin$ $\frac{3A}{2}$ $]$

$= 4 \cos$ $\frac{A}{2}$ $\sin$ $\frac{3A}{2}$ $\sin 3A =$ RHS. Hence proved.

iv) LHS $= \sin 3A + \sin 2A - \sin A$

$= [ \sin 3A - \sin A ] + \sin 2A$

$= 2 \sin \Big($ $\frac{3A- A}{2}$ $\Big) \cos \Big($ $\frac{3A+ A}{2}$ $\Big) + \sin 2A$

$= 2 \sin A \cos 2A + \sin 2A$

$= 2 \sin A \cos 2A + 2\sin A \cos A$

$= 2 \sin A [ \cos 2A + \cos A ]$

$= 2 \sin A \Big[ 2 \cos \Big($ $\frac{2A+ A}{2}$ $\Big) \cos \Big($ $\frac{2A- A}{2}$ $\Big) \Big]$

$= 4 \sin A \cos$ $\frac{3A}{2}$ $\cos$ $\frac{A}{2}$ $=$ RHS. Hence proved.

v) LHS $= \cos 20^o \cos 100^o + \cos 100^o \cos 140^o - \cos 140^o \cos 200^o$

$=$ $\frac{1}{2}$ $\Big[ 2\cos 20^o \cos 100^o + 2\cos 100^o \cos 140^o - 2\cos 140^o \cos 200^o \Big]$

$=$ $\frac{1}{2}$ $\Big[ \cos (100^o+20^o)+ \cos (100^o-20^o) + \cos (140^o+100^o)+ \\ \cos (140^o-100^o) - \cos (200^o+140^o) -\cos (200^o-140^o) \Big]$

$=$ $\frac{1}{2}$ $\Big[ \cos 120^o+ \cos 80^o + \cos 240^o+ \cos 40^o - \cos 340^o -\cos 60^o \Big]$

$=$ $\frac{1}{2}$ $\Big[ \cos (90^o+30^o)+ \cos 80^o + \cos (180^o+60^o)+ \cos 40^o - \cos (360^o-20^o) -\cos 60^o \Big]$

$=$ $\frac{1}{2}$ $\Big[ -\sin 30^o + [ \cos 80^o + \cos 40^o ] - \cos 60^o - \cos 20^o -\cos 60^o \Big]$

$=$ $\frac{1}{2}$ $\Big[ -\sin 30^o + 2 \cos 60^o \cos 20^o - \cos 60^o - \cos 20^o -\cos 60^o \Big]$

$=$ $\frac{1}{2}$ $\Big[ -$ $\frac{1}{2}$ $+ 2 \times$ $\frac{1}{2}$ $\cos 20^o -$ $\frac{1}{2}$ $- \cos 20^o -$ $\frac{1}{2}$ $\Big]$

$=$ $\frac{1}{2}$ $\Big[ -$ $\frac{3}{2}$ $\Big] = -$ $\frac{3}{4}$ $=$ RHS. Hence proved.

vi) LHS $= \sin$ $\frac{x}{2}$ $\sin$ $\frac{7x}{2}$ $+ \sin$ $\frac{3x}{2}$ $\sin$ $\frac{11x}{2}$

$=$ $\frac{1}{2}$ $\Big[ 2\sin$ $\frac{x}{2}$ $\sin$ $\frac{7x}{2}$ $+ 2\sin$ $\frac{3x}{2}$ $\sin$ $\frac{11x}{2}$ $\Big]$

$=$ $\frac{1}{2}$ $\Big[ \cos \Big($ $\frac{7x}{2}$ $-$ $\frac{x}{2}$ $\Big) - \cos \Big($ $\frac{7x}{2}$ $+$ $\frac{x}{2}$ $\Big) +\cos \Big($ $\frac{11x}{2}$ $-$ $\frac{3x}{2}$ $\Big) - \cos \Big($ $\frac{11x}{2}$ $+$ $\frac{3x}{2}$ $\Big) \Big]$

$=$ $\frac{1}{2}$ $\Big[ \cos 3x - \cos 4x + \cos 4x - \cos 7x \Big]$

$=$ $\frac{1}{2}$ $\Big[ \cos 3x - \cos 7x \Big]$

$= -$ $\frac{1}{2}$ $\Big[ \cos 7x - \cos 3x \Big]$

$= -$ $\frac{1}{2}$ $\Big[ -2 \sin \Big($ $\frac{7x+3x}{2}$ $\Big) \sin \Big($ $\frac{7x-3x}{2}$ $\Big) \Big]$

$= \sin 5x \sin 2x =$ RHS. Hence proved.

vii) LHS $= \cos x \cos$ $\frac{x}{2}$ $- \cos 3x \cos$ $\frac{9x}{2}$

$=$ $\frac{1}{2}$ $\Big[ 2\cos x \cos$ $\frac{x}{2}$ $- 2\cos 3x \cos$ $\frac{9x}{2}$ $\Big]$

$=$ $\frac{1}{2}$ $\Big[ \cos \Big( x +$ $\frac{x}{2}$ $\Big) + \cos \Big( x -$ $\frac{x}{2}$ $\Big) - \cos \Big($ $\frac{9x}{2}$ $+ 3x \Big) - \cos \Big($ $\frac{9x}{2}$ $- 3x \Big) \Big]$

$=$ $\frac{1}{2}$ $\Big[ \cos$ $\frac{3x}{2}$ $+ \cos$ $\frac{x}{2}$ $- \cos$ $\frac{15x}{2}$ $- \cos$ $\frac{3x}{2}$ $\Big]$

$=$ $\frac{1}{2}$ $\Big[ \cos$ $\frac{x}{2}$ $- \cos$ $\frac{15x}{2}$ $\Big]$

$=$ $\frac{1}{2}$ $\Big[ - 2 \sin \Big($ $\frac{\frac{x}{2}+\frac{15x}{2} }{2}$ $\Big) \sin \Big($ $\frac{\frac{x}{2}-\frac{15x}{2} }{2}$ $\Big) \Big]$

$= \sin 4x \sin$ $\frac{7x}{2}$ $=$ RHS. Hence proved.

$\\$

Question 7: Prove that:

i) $\frac{\sin A + \sin 3A}{\cos A - \cos 3A}$ $= \cot A$ ii) $\frac{\sin 9A - \sin 7A}{\cos 7A - \cos 9A}$ $= \cot 8A$

iii) $\frac{\sin A - \sin B}{\cos A + \cos B}$ $= \tan \Big($ $\frac{A-B}{2}$ $\Big)$ iv) $\frac{\sin A + \sin B}{\sin A - \sin B}$ $= \tan \Big($ $\frac{A+B}{2}$ $\Big) \cot \Big($ $\frac{A-B}{2}$ $\Big)$

v) $\frac{\cos A + \cos B}{\cos B - \cos A}$ $= \cot \Big($ $\frac{A+B}{2}$ $\Big) \cot \Big($ $\frac{A-B}{2}$ $\Big)$

Answer:

i) LHS $= \frac{\sin A + \sin 3A}{\cos A - \cos 3A}$

$=$ $\frac{2 \sin \Big( \frac{A+3A}{2} \Big) \cos \Big( \frac{A-3A}{2} \Big) }{-2 \sin \Big( \frac{A+3A}{2} \Big) \sin \Big( \frac{A-3A}{2} \Big)}$

$=$ $\frac{-2 \sin 2A \cos (-A) }{2 \sin 2A \sin (-A) }$ $=$ $\frac{- \cos ( -A)}{\sin ( -A)}$ $=$ $\frac{\cos A}{\sin A}$ $= \cot A =$ RHS. Hence proved.

ii) LHS $=$ $\frac{\sin 9A - \sin 7A}{\cos 7A - \cos 9A}$

$=$ $\frac{2 \sin \Big( \frac{9A-7A}{2} \Big) \cos \Big( \frac{9A+7A}{2} \Big) }{-2 \sin \Big( \frac{7A+9A}{2} \Big) \sin \Big( \frac{7A-9A}{2} \Big)}$

$=$ $\frac{-2 \sin A \cos 8A }{2 \sin 8A \sin (-A) }$ $=$ $\frac{- \sin A \cos 8A}{- \sin A \sin 8A }$ $=$ $\frac{\cos 8A}{\sin 8A}$ $= \cot 8A =$ RHS. Hence proved.

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

$=$ $\frac{2 \cos \Big( \frac{A+B}{2} \Big) \sin \Big( \frac{A-B}{2} \Big) }{2 \cos \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big)}$

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

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

$=$ $\frac{2 \sin \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big) }{2 \sin \Big( \frac{A-B}{2} \Big) \cos \Big( \frac{A+B}{2} \Big)}$

$= \tan \Big( \frac{A+B}{2} \Big) \cot \Big( \frac{A-B}{2} \Big) =$ RHS. Hence proved.

v) LHS $=$ $\frac{\cos A + \cos B}{\cos B - \cos A}$

$=$ $\frac{2 \cos \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big) }{-2 \sin \Big( \frac{B+A}{2} \Big) \sin \Big( \frac{B-A}{2} \Big)}$

$=$ $\frac{2 \cos \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big) }{2 \sin \Big( \frac{A+B}{2} \Big) \sin \Big( \frac{A-B}{2} \Big)}$

$= \cot \Big( \frac{A+B}{2} \Big) \cot \Big( \frac{A-B}{2} \Big) =$ RHS. Hence proved.

$\\$

Question 8: Prove that:

i) $\frac{\sin A + \sin 3A + \sin 5A}{\cos A + \cos 3A + \cos 5A}$ $= \tan 3A$

ii) $\frac{\cos 3A + 2\cos 5A + \cos 7A}{\cos A + 2\cos 3A + \cos 5A}$ $=$ $\frac{\cos 5A}{\cos 3A}$

iii) $\frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 2A}$ $= \cot 3A$

iv) $\frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A}$ $= \tan 6A$

v) $\frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{\cos 4A + \cos 7A - \cos 5A - \cos 8A}$ $= \cot 6A$

vi) $\frac{\sin 5A \cos 2A - \sin 6A \cos A}{\sin A \sin 2A - \cos 2A \cos 3A}$ $= \tan A$

vii) $\frac{\sin 11A \sin A + \sin 7A \sin 3A}{\cos 11A \sin A + \cos 7A \sin 3A}$ $= \tan 8A$

viii) $\frac{\sin 3A \cos 4A - \sin A \cos 2A}{\sin 4A \sin A + \cos 6A \cos A}$ $= \tan 2A$

ix) $\frac{\sin A \sin 2A + \sin 3A \sin 6A}{\sin A \cos 2A + \sin 3A \cos 6A}$ $= \tan 5A$

x) $\frac{\sin A + 2\sin 3A + \sin 5A}{\sin 3A + 2\sin 5A + \sin 7A}$ $=$ $\frac{\sin 3A}{\sin 5A}$

xi) $\frac{\sin (\theta + \phi) -2 \sin \theta + \sin (\theta - \phi)}{\cos (\theta + \phi) -2 \cos \theta + \cos (\theta - \phi)}$ $= \tan \theta$

Answer:

i) LHS $=$ $\frac{\sin A + \sin 3A + \sin 5A}{\cos A + \cos 3A + \cos 5A}$

$=$ $\frac{(\sin 5A + \sin A) + \sin 3A}{(\cos 5A + \cos A) + \cos 3A}$

$=$ $\frac{2 \sin \Big( \frac{5A+A}{2} \Big) \cos \Big( \frac{5A-A}{2} \Big) + \sin 3A } {2 \cos \Big( \frac{5A+A}{2} \Big) \cos \Big( \frac{5A-A}{2} \Big) + \cos 3A }$

$=$ $\frac{ 2 \sin 3A \cos 2A + \sin 3A}{ 2 \cos 3A \cos 2A+ \cos 3A}$

$=$ $\frac{ \sin 3A ( 2 \cos 2A + 1)}{\cos 3A ( 2 \cos 2A + 1)}$

$= \tan 3A =$ RHS. Hence proved.

ii) LHS $=$ $\frac{\cos 3A + 2\cos 5A + \cos 7A}{\cos A + 2\cos 3A + \cos 5A}$

$=$ $\frac{(\cos 3A + \cos 7A) + 2\cos 5A}{(\cos A + \cos 5A) + 2\cos 3A}$

$=$ $\frac{2 \cos \Big( \frac{7A+3A}{2} \Big) \cos \Big( \frac{7A-3A}{2} \Big) + 2\cos 5A } {2 \cos \Big( \frac{5A+A}{2} \Big) \cos \Big( \frac{5A-A}{2} \Big) + 2\cos 3A }$

$=$ $\frac{ 2 \cos 5A \cos 2A + 2\cos 5A}{ 2 \cos 3A \cos 2A+ 2\cos 3A}$

$=$ $\frac{ 2\cos 5A ( \cos 2A + 1)}{2\cos 3A ( \cos 2A + 1)}$

$=$ $\frac{\cos 5A}{\cos 3A}$ $=$ RHS. Hence proved.

iii) LHS $=$ $\frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 5A}$

$=$ $\frac{(\cos 4A + \cos 2A) + \cos 3A}{(\sin 4A + \sin 2A) + \sin 3A}$

$=$ $\frac{2 \cos \Big( \frac{4A+2A}{2} \Big) \cos \Big( \frac{4A-2A}{2} \Big) + \cos 3A } {2 \sin \Big( \frac{4A+2A}{2} \Big) \cos \Big( \frac{4A-2A}{2} \Big) + \sin 3A }$

$=$ $\frac{ 2 \cos 3A \cos A + \cos 3A}{ 2 \sin 3A \cos A+ \sin 3A}$

$=$ $\frac{ \cos 3A ( 2\cos A + 1)}{\sin 3A ( 2\cos A + 1)}$

$= \frac{\cos 3A}{\cos 3A}$ $= \cot 3A$ RHS. Hence proved.

iv) LHS $=$ $\frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A}$

$=$ $\frac{ (\sin 9A + \sin 3A) + (\sin 7A + \sin 5A) }{ (\cos 9A + \cos 3A) + (\cos 7A + \cos 5A) }$

$=$ $\frac{ 2 \sin \Big( \frac{9A+3A}{2} \Big) \cos \Big( \frac{9A-3A}{2} \Big) + 2 \sin \Big( \frac{7A+5A}{2} \Big) \cos \Big( \frac{7A-5A}{2} \Big) }{ 2 \cos \Big( \frac{9A+3A}{2} \Big) \cos \Big( \frac{9A-3A}{2} \Big) + 2 \cos \Big( \frac{7A+5A}{2} \Big) \cos \Big( \frac{7A-5A}{2} \Big) }$

$=$ $\frac{2 \sin 6A \cos 3A + 2 \sin 6A \cos 2A}{2 \cos 6A \cos 3A + 2 \cos 6A \cos 2A}$

$=$ $\frac{2 \sin 6A (\cos 3A + \cos 2A) }{2 \cos 6A (\cos 3A + \cos 2A)}$

$= \tan 6A =$ RHS. Hence proved.

v) LHS $=$ $\frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{ \cos 4A + \cos 7A - \cos 5A - \cos 8A}$

$=$ $\frac{ - (\sin 7A - \sin 5A) + (\sin 8A - \sin 4A) }{ - (\cos 7A - \cos 5A) - (\cos 8A - \cos 4A) }$

$=$ $\frac{ -2 \sin \Big( \frac{7A-5A}{2} \Big) \cos \Big( \frac{7A+5A}{2} \Big) + 2 \sin \Big( \frac{8A-4A}{2} \Big) \cos \Big( \frac{8A+4A}{2} \Big) }{- 2 \sin \Big( \frac{7A+5A}{2} \Big) \sin \Big( \frac{7A-5A}{2} \Big) +2 \sin \Big( \frac{8A+4A}{2} \Big) \sin \Big( \frac{8A-4A}{2} \Big) }$

$=$ $\frac{-2 \sin A \cos 6A + 2 \sin 2A \cos 6A}{-2 \sin 6A \sin A + 2 \sin 6A \sin 2A}$

$=$ $\frac{2 \cos 6A (-\sin A + \sin 2A) }{2 \sin 6A (-\sin A + \sin 2A) }$

$= \cot 6A =$ RHS. Hence proved.

vi) LHS $=$ $\frac{\sin 5A \cos 2A - \sin 6A \cos A}{\sin A \sin 2A - \cos 2A \cos 3A}$

$=$ $\frac{2\sin 5A \cos 2A - 2\sin 6A \cos A}{2\sin A \sin 2A - 2\cos 2A \cos 3A}$

$=$ $\frac{ \sin ( 5A +2A) + \sin (5A - 2A) - [ \sin ( 6A +A) + \sin (6A - A) ] }{ \cos ( 2A - A) - \cos ( 2A + A) - [ \cos ( 3A + 2A) + \cos ( 3A - 2 A) ] }$

$=$ $\frac{ \sin 7A + \sin 3A - [ \sin 7A + \sin 5A ] }{ \cos A - \cos 3A - [ \cos 5A + \cos A ] }$

$=$ $\frac{ \sin 3A - \sin 5A }{ - \cos 3A - \cos 5A }$

$=$ $\frac{ -( \sin 5A - \sin 3A) }{ - (\cos 5A + \cos 3A) }$

$=$ $\frac{2 \sin \Big( \frac{5A-3A}{2} \Big) \cos \Big( \frac{5A+3A}{2} \Big)}{2 \cos \Big( \frac{5A+3A}{2} \Big) \cos \Big( \frac{5A-3A}{2} \Big)}$

$=$ $\frac{\sin A \cos 4A}{\cos 4A \cos A}$

$=$ $\frac{\sin A }{\cos A}$ $= \tan A =$ RHS

vii) LHS $=$ $\frac{\sin 11A \sin A + \sin 7A \sin 3A}{\cos 11A \sin A + \cos 7A \sin 3A}$

$=$ $\frac{2\sin 11A \sin A + 2\sin 7A \sin 3A}{2\cos 11A \sin A + 2\cos 7A \sin 3A}$

$=$ $\frac{ \cos ( 11A -A) - \cos (11A + A) + \cos ( 7A -3A) - \cos (7A + 3A) }{ \sin ( 11A + A) - \sin ( 11A - A) + \sin ( 7A + 3A) - \sin ( 7A - 3 A) }$

$=$ $\frac{ \cos 10A - \cos 12A + \cos 4A - \cos 10A }{ \sin 12A - \sin 10A + \sin 10A - \sin 4A }$

$=$ $\frac{ - ( \cos 12A - \cos 4A) }{ \sin 12A - \sin 4A }$

$=$ $\frac{ - \Big[ -2 \sin \Big( \frac{12A+4A}{2} \Big) \cos \Big( \frac{12A-4A}{2} \Big) \Big] }{2 \sin \Big( \frac{12A-4A}{2} \Big) \cos \Big( \frac{12A+4A}{2} \Big)}$

$=$ $\frac{2\sin 8A \sin 4A}{2\sin 4A \cos 8A}$

$=$ $\frac{\sin 8A }{\cos 8A}$ $= \tan 8A =$ RHS

viii) LHS $=$ $\frac{\sin 3A \cos 4A - \sin A \cos 2A}{\sin 4A \sin A + \cos 6A \cos A}$

$=$ $\frac{2\sin 3A \cos 4A - 2\sin A \cos 2A}{2\sin 4A \sin A + 2\cos 6A \cos A}$

$=$ $\frac{ \sin ( 4A +3A) - \sin (4A -3A) - [ \sin ( 2A+A) - \sin (2A -A) ] }{ \cos ( 4A - A) - \cos ( 4A + A) + \cos ( 6A + A) + \cos ( 6A - A) }$

$=$ $\frac{ \sin 7A - \sin A - \sin 3A + \sin A }{ \cos 3A - \cos 5A + \cos 7A + \cos 5A }$

$=$ $\frac{ \sin 7A - \sin 3A }{ \cos 3A + \cos 7A }$

$=$ $\frac{2 \sin \Big( \frac{7A-3A}{2} \Big) \cos \Big( \frac{7A+3A}{2} \Big)}{2 \cos \Big( \frac{7A+3A}{2} \Big) \cos \Big( \frac{7A-3A}{2} \Big)}$

$=$ $\frac{2\sin 2A \cos 5A}{2\cos 5A \cos 2A}$

$=$ $\frac{\sin 2A }{\cos 2A}$ $= \tan 2A =$ RHS

ix) LHS $=$ $\frac{\sin A \sin 2A + \sin 3A \sin 6A}{\sin A \cos 2A + \sin 3A \cos 6A}$

$=$ $\frac{2\sin A \sin 2A + 2\sin 3A \sin 6A}{2\sin A \cos 2A + 2\sin 3A \cos 6A}$

$=$ $\frac{ \cos ( 2A -A) - \cos (2A +A) + \cos ( 6A-3A) - \cos (6A+3A) }{ \sin ( 2A + A) - \sin ( 2A - A) + \sin ( 6A + 3A) - \sin ( 6A - 3A) }$

$=$ $\frac{ \cos A - \cos 3A + \cos 3A - \cos 9A }{ \sin 3A - \sin A + \sin 9A - \sin 3A }$

$=$ $\frac{ \cos A - \cos 9A }{ - \sin A + \sin 9A }$

$=$ $\frac{ - (\cos 9A - \cos A) }{ ( \sin 9A - \sin A) }$

$=$ $\frac{- \Big[ -2 \sin \Big( \frac{9A+A}{2} \Big) \sin \Big( \frac{9A-A}{2} \Big) \Big] }{2 \sin \Big( \frac{9A-A}{2} \Big) \cos \Big( \frac{9A+A}{2} \Big)}$

$=$ $\frac{2\sin 5A \sin 4A}{2\sin 4A \cos 5A}$

$=$ $\frac{\sin 5A }{\cos 5A}$ $= \tan 5A =$ RHS

x) LHS $=$ $\frac{\sin A + 2\sin 3A + \sin 5A}{\sin 3A + 2\sin 5A + \sin 7A}$

$=$ $\frac{ (\sin A + \sin 5A )+ 2\sin 3A }{ (\sin 3A + \sin 7A) + 2\sin 5A }$

$=$ $\frac{2 \sin \Big( \frac{5A+A}{2} \Big) \cos \Big( \frac{5A-A}{2} \Big) + 2\sin 3A } {2 \sin \Big( \frac{7A+3A}{2} \Big) \cos \Big( \frac{7A-3A}{2} \Big) + 2\sin 5A }$

$=$ $\frac{ 2 \sin 3A \cos 2A + 2 \sin 3A}{ 2 \sin 5A \cos 2A+ 2\sin 5A}$

$=$ $\frac{ \sin 3A ( 1+ 2\cos A )}{\sin 5A ( 1+ 2\cos A )}$

$= \frac{\sin 3A}{\sin 5A}$ $= \cot 3A$ RHS. Hence proved.

xi) LHS $=$ $\frac{\sin (\theta + \phi) -2 \sin \theta + \sin (\theta - \phi)}{\cos (\theta + \phi) -2 \cos \theta + \cos (\theta - \phi)}$ $= \tan \theta$

$=$ $\frac{ [ \sin (\theta + \phi) + \sin (\theta - \phi) ] -2 \sin \theta}{ [ \cos (\theta + \phi) + \cos (\theta - \phi) ] -2 \cos \theta }$

$=$ $\frac{ 2 \sin \Big( \frac{\theta + \phi+\theta - \phi}{2} \Big) \cos \Big( \frac{\theta + \phi-\theta + \phi}{2} \Big) -2 \sin \theta}{ 2 \cos \Big( \frac{\theta + \phi+\theta - \phi}{2} \Big) \cos \Big( \frac{\theta + \phi-\theta + \phi}{2} \Big) -2 \cos \theta }$

$=$ $\frac{2\sin \theta \cos \phi - 2 \sin \theta}{2\cos \theta \cos \phi - 2 \cos \theta}$

$=$ $\frac{2\sin \theta ( \cos \phi - 1) }{2\cos \theta ( \cos \phi - 1)}$

$=$ $\frac{\sin \theta }{\cos \theta}$

$= \tan \theta =$ RHS. Hence proved.

$\\$

Question 9: Prove that:

i) $\sin \alpha + \sin \beta + \sin \gamma - \sin (\alpha + \beta + \gamma) = 4 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big) \sin \Big($ $\frac{\beta + \gamma}{2}$ $\Big) \sin \Big($ $\frac{\gamma+ \alpha}{2}$ $\Big)$

ii) $\cos ( A+B+C) + \cos (A-B+C) + \cos ( A + B - C) + \cos (-A+B+C) = 4 \cos A \cos B \cos C$

Answer:

i) LHS $= \sin \alpha + \sin \beta + \sin \gamma - \sin (\alpha + \beta + \gamma)$

$= [ \sin \alpha + \sin \beta ] + [ \sin \gamma - \sin (\alpha + \beta + \gamma) ]$

$= 2 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big) \cos \Big($ $\frac{\alpha - \beta}{2}$ $\Big) + 2 \sin \Big($ $\frac{\gamma - (\alpha + \beta + \gamma)}{2}$ $\Big) \cos \Big($ $\frac{\gamma + (\alpha + \beta + \gamma)}{2}$ $\Big)$

$= 2 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big) \cos \Big($ $\frac{\alpha - \beta}{2}$ $\Big) + 2 \sin \Big($ $\frac{ - (\alpha + \beta )}{2}$ $\Big) \cos \Big($ $\frac{\alpha + \beta + 2\gamma}{2}$ $\Big)$

$= 2 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big) \Big[ \cos \Big($ $\frac{\alpha - \beta}{2}$ $\Big) - \cos \Big($ $\frac{\alpha + \beta + 2\gamma)}{2}$ $\Big) \Big]$

$= 2 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big) \Big[ - 2 \sin \Big($ $\frac{\frac{\alpha - \beta}{2} +\frac{\alpha + \beta + 2\gamma}{2}}{2}$ $\Big) \sin \Big($ $\frac{\frac{\alpha - \beta}{2} - \frac{\alpha + \beta + 2\gamma}{2}}{2}$ $\Big) \Big]$

$= 2 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big)\Big[ -2 \sin \Big($ $\frac{\alpha + \gamma}{2}$ $\Big) \sin \Big($ $\frac{-(\beta + \gamma)}{2}$ $\Big) \Big]$

$= 4 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big)\Big[ \sin \Big($ $\frac{\alpha + \gamma}{2}$ $\Big) \sin \Big($ $\frac{\beta + \gamma}{2}$ $\Big) \Big]$

$= 4 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big)\Big[ \sin \Big($ $\frac{\beta + \gamma}{2}$ $\Big) \sin \Big($ $\frac{\alpha + \gamma}{2}$ $\Big) \Big] =$ RHS. Hence proved.

ii) LHS $= \cos ( A+B+C) + \cos (A-B+C) + \cos ( A + B - C) + \cos (-A+B+C)$

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

$= 2 \cos \Big( \frac{A+B+C + A - B + C}{2} \cos \frac{A + B + C - A + B - C}{2} \Big) + 2 \cos \Big( \frac{A+B-C - A + B + C}{2} \cos \frac{A + B - C + A - B - C}{2} \Big)$

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

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

$= 2 \cos B \Big[ 2 \cos \Big($ $\frac{A + C + A - C}{2}$ $\Big) \cos \Big($ $\frac{A + C - A +C}{2}$ $\Big) \Big]$

$= 4 \cos B \cos A \cos C$

$= 4 \cos A \cos B \cos C =$ RHS. Hence proved.

$\\$

Question 10: If $\cos A + \cos B =$ $\frac{1}{2}$ and $\sin A + \sin B =$ $\frac{1}{4}$ , prove that:

$\tan \Big($ $\frac{A+B}{2}$ $\Big) =$ $\frac{1}{2}$

Answer:

Given: $\cos A + \cos B =$ $\frac{1}{2}$ and $\sin A + \sin B =$ $\frac{1}{4}$

Dividing one by another

$\Rightarrow$ $\frac{\sin A + \sin B}{\cos A + \cos B}$ $=$ $\frac{\frac{1}{4}}{\frac{1}{2}}$

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

$\Rightarrow$ $\frac{ \sin \Big( \frac{A+B}{2} \Big) }{ \cos \Big( \frac{A+B}{2} \Big) }$ $=$ $\frac{1}{2}$

$\Rightarrow \tan \Big($ $\frac{A+B}{2}$ $\Big) =$ $\frac{1}{2}$

$\\$

Question 11: If $\mathrm{cosec} A + \sec A = \mathrm{cosec} B + \sec B$ , then prove that $\tan A \tan B = \cot \Big($ $\frac{A+B}{2}$ $\Big)$

Answer:

Given: $\mathrm{cosec} A + \sec A = \mathrm{cosec} B + \sec B$

$\Rightarrow \sec B - \sec A = \mathrm{cosec} B - \mathrm{cosec} A$

$\Rightarrow$ $\frac{1}{\cos A}$ $-$ $\frac{1}{\cos B}$ $=$ $\frac{1}{\sin B}$ $- \frac{1}{\sin A}$

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

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

$\Rightarrow \tan A \tan B =$ $\frac{2 \sin \frac{A-B}{2} \cos \frac{A+B}{2}}{2 \sin \frac{B-A}{2} \sin \frac{B+A}{2}}$

$\Rightarrow \tan A \tan B = \cot$ $\frac{A+B}{2}$

$\\$

Question 12: If $\sin 2A = \lambda \sin 2B$ , prove that $\frac{\tan (A+B)}{\tan (A-B)}$ $=$ $\frac{\lambda+1}{\lambda - 1}$

Answer:

Given $\sin 2A = \lambda \sin 2B$

$\Rightarrow \lambda =$ $\frac{\sin 2A}{\sin 2B}$

Applying componendo and dividendo

$\frac{\lambda+1}{\lambda-1}$ $=$ $\frac{\sin 2A + \sin 2B}{\sin 2A - \sin 2B}$

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

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

Hence proved.

$\\$

Question 13: Prove that:

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

ii) $\sin (B-C) \cos (A-D) + \sin (C-A) \cos (B-D) + \sin (A-B) \cos (C-D) = 0$

Answer:

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

$=$ $\frac{2 \cos \Big( \frac{A+B+C-A+B+C}{2} \Big) \cos \Big( \frac{A+B+C+A-B-C}{2} \Big) + 2 \cos \Big( \frac{A-B+C+A+B-C}{2} \Big) \cos \Big( \frac{A-B+C-A-B+C}{2} \Big) }{2 \sin \Big( \frac{A+B+C-A+B+C}{2} \Big) \cos \Big( \frac{A+B+C+A-B-C}{2} \Big) + 2 \sin \Big( \frac{A-B+C+A+B-C}{2} \Big) \cos \Big( \frac{A-B+C-A-B+C}{2} \Big) }$

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

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

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

$=$ $\frac{ 2 \cos \Big( \frac{B+C+C-B}{2} \Big) \cos \Big( \frac{B+C-C+B}{2} \Big) }{ 2 \sin \Big( \frac{B+C+C-B}{2} \Big) \cos \Big( \frac{B+C-C+B}{2} \Big) }$

$=$ $\frac{2 \cos C \cos B}{ 2 \sin C \cos B}$

$=$ $\frac{\cos C}{\sin C}$ $= \cot C$

ii) LHS $= \sin (B-C) \cos (A-D) + \sin (C-A) \cos (B-D) + \sin (A-B) \cos (C-D)$

$=$ $\frac{1}{2}$ $[ 2\sin (B-C) \cos (A-D) + 2\sin (C-A) \cos (B-D) + 2\sin (A-B) \cos (C-D) ]$

$=$ $\frac{1}{2}$ $[ \sin ( B-C+A-D ) + \sin ( B - C - A + D) + \sin ( C - A + B - D) + \\ \sin ( C - A - B + D) + \sin (A - B + C - D) + \sin ( A - B - C + D) ]$

$=$ $\frac{1}{2}$ $[ \sin ( A + B - C - D ) + \sin ( B+D - A - C) + \sin ( B+C - A - D) \\ - \sin ( A + B - C - D ) - \sin (B + D - A - C) - \sin ( B+C-A-D) ]$

$=$ $\frac{1}{2}$ $[ 0 ] =$ RHS. Hence proved.

$\\$

Question 14: If $\frac{\cos (A-B) }{\cos (A+B)}$ $+$ $\frac{\cos (C+D)}{\cos (C-D)}$ $= 0$ , prove that $\tan A \tan B \tan C \tan D = -1$

Answer:

Given $\frac{\cos (A-B) }{\cos (A+B)}$ $+$ $\frac{\cos (C+D)}{\cos (C-D)}$ $= 0$

$\Rightarrow$ $\frac{\cos ( A - B) \cos ( C - D) + \cos ( C+D) \cos (A+B) }{\cos ( A+ B) \cos ( C+D)}$ $= 0$

$\Rightarrow \cos ( A - B) \cos ( C - D) + \cos ( C+D) \cos (A+B) = 0$

$\Rightarrow \cos ( A - B) \cos ( C - D) = - \cos ( C+D) \cos (A+B)$

$\Rightarrow (\cos A \cos B + \sin A \sin B) (\cos C \cos D + \sin C \sin D) = - ( \cos C \cos D - \sin C \sin D)(\cos A \cos B - \sin A \sin B)$

Dividing both sides by $\cos A \cos B \cos C \cos D$

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

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

$\Rightarrow ( 1 + \tan A \tan B) ( 1 + \tan C \tan D) = - ( 1- \tan C \tan D) ( 1 - \tan A \tan B)$

$\Rightarrow ( 1 + \tan A \tan B) ( 1 + \tan C \tan D) = ( \tan C \tan D - 1) ( 1 - \tan A \tan B)$

$\Rightarrow 1 + \tan A \tan B + \tan C \tan D + \tan A \tan B \tan C \tan D = \tan C \tan D - \tan A \tan B \tan C \tan D - 1 + \tan A \tan B$

$\Rightarrow 2 \tan A \tan B \tan C \tan D = -2$

$\Rightarrow \tan A \tan B \tan C \tan D = -1$

Hence proved.

$\\$

Question 15: If $\cos (\alpha + \beta) \sin (\gamma + \delta) = \cos (\alpha - \beta) \sin (\gamma - \delta)$ , then prove that $\cot \alpha \cot \beta \cot \gamma = \cot \delta$

Answer:

Given $\cos (\alpha + \beta) \sin (\gamma + \delta) = \cos (\alpha - \beta) \sin (\gamma - \delta)$

$\Rightarrow [ \cos \alpha \cos \beta - \sin \alpha \sin \beta ] [ \sin \gamma \cos \delta + \cos \gamma \sin \delta ] = [ \cos \alpha \cos \beta + \sin \alpha \sin \beta ] [ \sin \gamma \cos \delta - \cos \gamma \sin \delta ]$

Dividing both sides by $\sin \alpha \sin \beta \sin \gamma \sin \delta$

$\Rightarrow$ $\frac{ [ \cos \alpha \cos \beta - \sin \alpha \sin \beta ] [ \sin \gamma \cos \delta + \cos \gamma \sin \delta ]}{\sin \alpha \sin \beta \sin \gamma \sin \delta}$ $=$ $\frac{ [ \cos \alpha \cos \beta + \sin \alpha \sin \beta ] [ \sin \gamma \cos \delta - \cos \gamma \sin \delta ]}{\sin \alpha \sin \beta \sin \gamma \sin \delta}$

$\Rightarrow (\cot \alpha \cot \beta - 1 ) ( \cot \delta + \cot \gamma) = (\cot \alpha \cot \beta + 1 ) ( \cot \delta - \cot \gamma)$

$\Rightarrow \cot \alpha \cot \beta \cot \delta - \cot \delta + \cot \alpha \cot \beta \cot \gamma - \cot \gamma = \cot \alpha \cot \beta \cot \delta + \cot \delta - \cot \alpha \cot \beta \cot \gamma - \cot \gamma$

$\Rightarrow 2 \cot \alpha \cot \beta \cot \gamma = 2 \cot \delta$

$\Rightarrow \cot \alpha \cot \beta \cot \gamma = \cot \delta$

Hence proved.

$\\$

Question 16: If $y \sin \phi = x \sin (2 \theta + \phi)$ , prove that $(x+y) \cot (\theta + \phi)= (y-x) \cot \theta$

Answer:

Given $y \sin \phi = x \sin (2 \theta + \phi)$

$\Rightarrow$ $\frac{y}{x}$ $=$ $\frac{\sin (2 \theta + \phi)}{\sin \phi}$

Applying componendo and dividendo

$\frac{y-x}{y+x}$ $=$ $\frac{\sin (2 \theta + \phi) - \sin \phi}{ \sin (2 \theta + \phi) + \sin \phi}$

$\Rightarrow$ $\frac{y-x}{y+x}$ $=$ $\frac{2 \sin \Big( \frac{2 \theta + \phi - \phi}{2} \Big) \cos \Big( \frac{2 \theta + \phi + \phi}{2} \Big)}{ 2 \sin \Big( \frac{2 \theta + \phi + \phi}{2} \Big) \cos \Big( \frac{2 \theta + \phi - \phi}{2} \Big)}$

$\Rightarrow$ $\frac{y-x}{y+x}$ $=$ $\frac{\sin \theta \cos ( \theta + \phi) }{\sin ( \theta + \phi) \cos \theta}$

$\Rightarrow$ $\frac{y-x}{y+x}$ $=$ $\frac{\cot ( \theta + \phi) }{\cos \theta}$

$\Rightarrow (y-x) \cos \theta = (y+x) \cot ( \theta + \phi)$

Hence proved.

$\\$

Question 17: If $\cos (A+B) \sin (C-D) = \cos (A-B) \sin (C+D)$ , prove that $\tan A \tan B \tan C + \tan D = 0$

Answer:

Given $\cos (A+B) \sin (C-D) = \cos (A-B) \sin (C+D)$

$\Rightarrow (\cos A \cos B - \sin A \sin B) ( \sin C \cos D - \cos C \sin D) = (\cos A \cos B + \sin A \sin B) ( \sin C \cos D + \cos C \sin D)$

Divide both sides by $\cos A \cos B \cos C\cos D$ we get

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

$\Rightarrow ( 1 - \tan A \tan B)(\tan C - \tan D) = (1 + \tan A \tan B) (\tan C + \tan D)$

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

$\Rightarrow -2 \tan D = 2 \tan A \tan B \tan C$

$\Rightarrow \tan A \tan B \tan C + \tan D = 0$

Hence proved.

$\\$

Question 18: If $x \cos \theta = y \cos \Big( \theta+$ $\frac{2\pi}{3}$ $\Big) = z \cos \Big( \theta +$ $\frac{4\pi}{3}$ $\Big)$ , prove that $xy + yz + zx = 0$

Answer:

Given $x \cos \theta = y \cos \Big( \theta+$ $\frac{2\pi}{3}$ $\Big) = z \cos \Big( \theta +$ $\frac{4\pi}{3}$ $\Big) = k$

$\Rightarrow x =$ $\frac{k}{\cos \theta}$

$\Rightarrow y =$ $\frac{k}{\cos \Big( \theta + \frac{2\pi}{3} \Big) }$

$\Rightarrow z =$ $\frac{k}{\cos \Big( \theta + \frac{4\pi}{3} \Big) }$

$\therefore xy + yz + zx$

$= \Big[$ $\frac{k}{\cos \theta} \frac{k}{\cos \big( \theta + \frac{2\pi}{3} \big) } + \frac{k}{\cos \big( \theta + \frac{2\pi}{3} \big) } \frac{k}{\cos \big( \theta + \frac{4\pi}{3} \big) } + \frac{k}{\cos \big( \theta + \frac{4\pi}{3} \big) } \frac{k}{\cos \theta}$ $\Big]$

$= k^2 \Big[$ $\frac{1}{\cos \theta} \frac{1}{\cos \big( \theta + \frac{2\pi}{3} \big) } + \frac{1}{\cos \big( \theta + \frac{2\pi}{3} \big) } \frac{1}{\cos \big( \theta + \frac{4\pi}{3} \big) } + \frac{1}{\cos \big( \theta + \frac{4\pi}{3} \big) } \frac{1}{\cos \theta}$ $\Big]$

$= k^2 \Big[$ $\frac{\cos \big( \theta + \frac{4\pi}{3} \big) + \cos \theta + \cos \big( \theta + \frac{2\pi}{3} \big)}{\cos \big( \theta+ \frac{4\pi}{3} \big) \cos \theta \cos \big( \theta + \frac{2\pi}{3} \big)}$ $\Big]$

$= k^2 \Big[$ $\frac{ \cos \theta \cos \frac{4\pi}{3} - \sin \theta \sin \frac{4\pi}{3} + \cos \theta + \cos \theta \cos \frac{2\pi}{3} - \sin \theta \sin \frac{2\pi}{3}}{\cos \big( \theta + \frac{4\pi}{3} \big) \cos \theta \cos \big( \theta + \frac{2\pi}{3} \big)}$ $\Big]$

$= k^2 \Big[$ $\frac{ \cos \theta (-\frac{1}{2}) - \sin \theta (-\frac{\sqrt{3}}{2}) + \cos \theta + \cos \theta (-\frac{1}{2}) - \sin \theta (\frac{\sqrt{3}}{2}) }{\cos \big( \theta + \frac{4\pi}{3} \big) \cos \theta \cos \big( \theta + \frac{2\pi}{3} \big)}$ $\Big]$

$= k^2 \Big[$ $\frac{ -\cos \theta + (\frac{\sqrt{3}}{2}) \sin \theta + \cos \theta - (\frac{\sqrt{3}}{2}) \sin \theta }{\cos \big( \theta + \frac{4\pi}{3} \big) \cos \theta \cos \big( \theta + \frac{2\pi}{3} \big)}$ $\Big]$

$= k^2 \Big[$ $\frac{ 0 }{\cos \big( \theta + \frac{4\pi}{3} \big) \cos \theta \cos \big( \theta + \frac{2\pi}{3} \big)}$ $\Big]$

$= 0 =$ RHS. Hence proved.

$\\$

Question 19: If $m \sin \theta = n \sin (\theta + 2 \alpha)$ , prove that $\tan (\theta + \alpha) \cot \alpha =$ $\frac{m+n}{m-n}$

Answer:

Given: $m \sin \theta = n \sin (\theta + 2 \alpha)$

$\frac{m}{n}$ $=$ $\frac{\sin (\theta + 2 \alpha)}{\sin \theta}$

Applying componendo and dividendo

$\frac{m+n}{m-n}$ $=$ $\frac{\sin (\theta + 2 \alpha) +\sin \theta }{\sin (\theta + 2 \alpha) -\sin \theta}$

$\frac{m+n}{m-n}$ $=$ $\frac{2 \sin \big( \frac{\theta + 2 \alpha + \theta}{2} \big) \cos \big( \frac{\theta + 2 \alpha - \theta}{2} \big) }{2 \cos \big( \frac{\theta + 2 \alpha + \theta}{2} \big) \sin \big( \frac{\theta + 2 \alpha - \theta}{2} \big)}$