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)}

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

\frac{m+n}{m-n} = \frac{\tan ( \theta + \alpha) }{\tan \alpha}

\tan ( \theta + \alpha) = \frac{m+n}{m-n} \tan \alpha

Hence proved.