Prove the following identities (1-25)

Question 1: \sqrt{\frac{1 - \cos 2x}{1+\cos 2x} } = \tan x

Answer:

LHS = \sqrt{\frac{1 - \cos 2x}{1+\cos 2x} } = \sqrt{\frac{2 \sin^2 x}{2 \cos^2 x} } = \frac{\sin x}{\cos x} = \tan x = RHS. Hence proved.

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Question 2: \frac{\sin 2x}{1- \cos 2x} = \cot x

Answer:

LHS = \frac{\sin 2x}{1- \cos 2x} = \frac{2 \sin x \cos x}{2 \sin^2 x} = \frac{\cos x}{\sin x} = \cot x = RHS. Hence proved.

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Question 3: \frac{\sin 2x}{1+ \cos 2x} = \tan x

Answer:

LHS = \frac{\sin 2x}{1+ \cos 2x} = \frac{2 \sin x \cos x}{2 \cos^2 x} = \frac{\sin x}{\cos x} = \tan x = RHS. Hence proved.

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Question 4: \sqrt{2 + \sqrt{2 + 2 \cos 4x} } = 2 \cos x, 0 < x < \frac{\pi}{4}

Answer:

LHS = \sqrt{2 + \sqrt{2 + 2 \cos 4x} }

= \sqrt{2 + \sqrt{2(1 +  \cos 4x)} }

= \sqrt{2 + \sqrt{2 \times 2 \cos^2 2x} }

= \sqrt{2 + 2 \cos 2x }

= \sqrt{2(1 +  \cos 2x) }

= \sqrt{2 \times 2 \cos^2 x }

= 2 \cos x = RHS. Hence proved.

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Question 5: \frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x} = \tan x

Answer:

LHS = \frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x}

= \frac{2\sin^2 x + 2 \sin x \cos x}{2 \cos^2 x+ 2 \sin x \cos x}

= \frac{2\sin x (\sin x + \cos x)}{2\cos x (\sin x + \cos x)}

= \tan x = RHS. Hence proved.

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Question 6: \frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} = \tan x

Answer:

LHS = \frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x}

= \frac{\sin x + 2 \sin x \cos x }{\cos x + (1 + \cos 2x) }

= \frac{\sin x (1 + 2 \cos x)  }{\cos x + 2 \cos^2 x }

= \frac{\sin x (1 + 2 \cos x)  }{\cos x ( 1  + 2 \cos x) }

= \frac{\sin x   }{\cos x }

= \tan x =   RHS. Hence proved.

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Question 7: \frac{\cos 2x}{1+ \sin 2x} = \tan \Big( \frac{\pi}{4}   - x \Big)

Answer:

LHS = \frac{\cos 2x}{1+ \sin 2x}

= \frac{\cos^2 x - \sin^2 x}{\sin^2 x + \cos^2 x + 2 \sin x \cos x }

= \frac{(\cos x - \sin x ) ( \cos x + \sin x) }{( \cos x + \sin x)^2}

= \frac{\cos x - \sin x   }{\cos x + \sin x}

= \frac{1 - \tan x}{1+ \tan x}

= \frac{\tan \frac{\pi}{4} - \tan x}{1+ \tan \frac{\pi}{4} \tan x}

= \tan \Big( \frac{\pi}{4}   - x \Big) RHS. Hence proved.

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Question 8: \frac{\cos x}{1- \sin x} = \tan \Big( \frac{\pi}{4}   + \frac{x}{2} \Big)

Answer:

LHS = \frac{\cos x}{1 - \sin x}

= \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{\sin^2 \frac{x}{2}+ \cos^2 \frac{x}{2} -  2 \sin \frac{x}{2} \cos \frac{x}{2} }

= \frac{(\cos \frac{x}{2} - \sin \frac{x}{2} ) ( \cos \frac{x}{2} + \sin \frac{x}{2}) }{( \cos \frac{x}{2} - \sin \frac{x}{2})^2}

= \frac{\cos \frac{x}{2} + \sin \frac{x}{2}   }{\cos \frac{x}{2} - \sin \frac{x}{2}}

= \frac{1 + \tan \frac{x}{2}}{1- \tan \frac{x}{2}}

= \frac{\tan \frac{\pi}{4} + \tan \frac{x}{2}}{1- \tan \frac{\pi}{4} \tan \frac{x}{2}}

= \tan \Big( \frac{\pi}{4} + \frac{x}{2} \Big) RHS. Hence proved.

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Question 9: \cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8} +\cos^2 \frac{5\pi}{8} +\cos^2 \frac{7\pi}{8} = 2

Answer:

LHS = \cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8} +\cos^2 \frac{5\pi}{8} +\cos^2 \frac{7\pi}{8}

= \cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8} +\cos^2 \big( \pi - \frac{3\pi}{8}  \big) +\cos^2\big( \pi - \frac{\pi}{8}  \big)

= \cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8} +\cos^2 \frac{3\pi}{8} +\cos^2 \frac{\pi}{8}

= 2 \Big( \cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8} \Big)

= 2 \Big( \cos^2 \frac{\pi}{8} + \cos^2 \big( \frac{\pi}{2} - \frac{\pi}{8}  \big) \Big)

= 2 \Big( \cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8} \Big)

= 2 = RHS. Hence proved.

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Question 10: \sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8} + \sin^2 \frac{5\pi}{8} +\sin^2 \frac{7\pi}{8} = 2

Answer:

LHS = \sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8} + \sin^2 \frac{5\pi}{8} +\sin^2 \frac{7\pi}{8}

= \sin^2 \frac{\pi}{8} + \sin^2 \big( \frac{\pi}{2} - \frac{\pi}{8} \big) + \sin^2 \big( \pi - \frac{3\pi}{8} \big) +\sin^2 \big( \frac{\pi}{2} - \frac{\pi}{8} \big)

= \sin^2 \frac{\pi}{8} + \cos^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8} +\sin^2 \frac{\pi}{8}

= 1+ \sin^2 \Big( \frac{\pi}{2} - \frac{\pi}{8} \Big) + \sin^2 \frac{\pi}{8}

= 1+ \cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8}

= 2 = RHS. Hence proved.

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Question 11: (\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2 = 4 \cos^2 \Big( \frac{\alpha - \beta}{2} \Big)

Answer:

LHS = (\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2

= \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta + \sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta

= 2 + 2 \cos \alpha \cos \beta  + 2 \sin \alpha \sin \beta

= 2 + 2 \cos ( \alpha - \beta)

= 2 [ 1 + \cos ( \alpha - \beta) ]

= 2 \times 2 \cos^2 \Big( \frac{\alpha - \beta}{2} \Big)

= 4 \cos^2 \Big( \frac{\alpha - \beta}{2} \Big) = RHS. Hence proved.

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Question 12: \sin^2 \Big( \frac{\pi}{8} + \frac{x}{2} \Big) - \sin^2 \Big( \frac{\pi}{8} - \frac{x}{2} \Big) = \frac{1}{\sqrt{2}} \sin x

Answer:

LHS = \sin^2 \Big( \frac{\pi}{8} + \frac{x}{2} \Big) - \sin^2 \Big( \frac{\pi}{8} - \frac{x}{2} \Big)

= \frac{1}{2} \Big[ 1 - \cos 2 \Big( \frac{\pi}{8} + \frac{x}{2} \Big) \Big]  - \frac{1}{2} \Big[ 1 - \cos  2\Big( \frac{\pi}{8} - \frac{x}{2} \Big) \Big]

= \frac{1}{2} \Big[ \cos  2\Big( \frac{\pi}{8} - \frac{x}{2} \Big) - \cos 2 \Big( \frac{\pi}{8} + \frac{x}{2} \Big) \Big]   

= \frac{1}{2} \Big[ \cos  \Big( \frac{\pi}{4} -  x  \Big) - \cos  \Big( \frac{\pi}{4} + x \Big) \Big]   

= \frac{1}{2} \Big[ -2 \sin \Big( \frac{\frac{\pi}{4} - x \frac{\pi}{4}+x}{2} \Big) \sin \Big( \frac{\frac{\pi}{4}-x-\frac{\pi}{4}-x}{2} \Big) \Big]

= \frac{1}{2} \Big[ - 2 \sin \frac{\pi}{4} \sin ( -x) \Big]

= \frac{1}{\sqrt{2}} \sin x = RHS. Hence proved.

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Question 13: 1 + \cos^2 2x = 2 (\cos^4 x + \sin^4 x)

Answer:

LHS = 1 + \cos^2 2x

= 1 + ( \cos^2 x - \sin^2 x)^2

= 1 + \cos^4 x + \sin^4 x - 2 \cos^2 x \sin^2 x

= ( \cos^2 x + \sin^2 x)^2 + \cos^4 x + \sin^4 x - 2 \cos^2 x \sin^2 x

= \cos^4 x + \sin^4 x + 2 \cos^2 x \sin^2 x + \cos^4 x + \sin^4 x - 2 \cos^2 x \sin^2 x

= 2 ( \cos^4 x + \sin^4 x) = RHS. Hence proved.

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Question 14: \cos^3 2x + 3 \cos 2x = 4 ( \cos^6 x - \sin^6 x)

Answer:

RHS = 4 ( \cos^6 x - \sin^6 x)

= 4 [ (\cos^2 x)^3 - (\sin^2 x)^3 ]

= 4 [ (\cos^2 x - \sin^2 x   ) ( \cos^4 x  + \sin^4 x + \sin^2 x \cos^2 x )]

= 4 \cos 2x ( \cos^4 x  + \sin^4 x + \sin^2 x \cos^2 x )

= 4 \cos 2x [ (\cos^2 x  - \sin^2 x)^2 + 2 \cos^2 x \sin^2 x+ \sin^2 x \cos^2 x ]

= 4 \cos 2x [ \cos^2 2x + 3 \sin^2 x \cos^2 x ]

= 4 \cos 2x \Big[ \cos^2 2x + 3 \Big( \frac{1 - \cos 2x}{2} \Big) \Big( \frac{1 + \cos 2x}{2} \Big) \Big ]

= 4 \cos 2x \Big[ \cos^2 2x + \frac{3}{4} (1 - \cos^2 2x) \Big]

= \cos 2x [ 4\cos^2 2x + 3 (1 - \cos^2 2x) ]

= \cos 2x [ \cos^2 2x + 3 ]

= \cos^3 2x + 3 \cos 2x = LHS. Hence proved.

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Question 15: (\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x = 0

Answer:

LHS = (\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x

= 2 \sin 2x \cos x \sin x - 2 \sin 2x \sin x \cos x

= 2 \sin 2x \cos x \sin x - 2 \sin 2x  \cos x \sin x

= 0 = RHS. Hence proved.

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Question 16: \cos^2 \Big( \frac{\pi}{4} -  x  \Big) - \sin^2 \Big( \frac{\pi}{4} - x \Big) = \sin 2x

Answer:

LHS = \cos^2 \Big( \frac{\pi}{4} -  x  \Big) - \sin^2 \Big( \frac{\pi}{4} - x \Big)

Since \cos 2x = \cos^2 x - \sin^2 x

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

= \cos  ( \frac{\pi}{2} - 2x)

= \sin 2x = RHS. Hence proved.

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Question 17: \cos 4x = 1 - 8 \cos^2 x + 8 \cos ^4 x

Answer:

LHS = \cos 4x

= \cos 2 \times 2x

= 2 \cos^2 2x - 1

= 2 [ ( 2 \cos^2 x - 1)^2 ] - 1

= 2 ( 4 \cos^4 x + 1 - 4 \cos^2 x ) - 1

= 8 \cos^4 x - 8 \cos^2 x + 1

= RHS. Hence proved.

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Question 18: \sin 4x = 4 \sin x \cos^3 x - 4 \cos x \sin^3 x

Answer:

LHS = \sin 4x

= \sin 2 .2x

= 2 \sin 2x \cos 2x

= 2 ( 2 \sin x \cos x)(\cos^2 x - \sin^2 x)

= 4 \sin x \cos^3 x - 4 \sin^3 x \cos x

= RHS. Hence proved.

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Question 19: 3(\sin x - \cos x)^4 + 6 (\sin x + \cos x)^2 + 4 (\sin^6 x + \cos^6 x) = 13

Answer:

LHS = 3(\sin x - \cos x)^4 + 6 (\sin x + \cos x)^2 + 4 (\sin^6 x + \cos^6 x)

= 3 ( \sin^4 x - 4 \sin^3x \cos x + 6 \sin^2 x \cos^2 x - 4 \sin x \cos^3 x + \cos^4 x) + 6 ( \sin^2 x + 2 \sin x \cos x + \cos^2 x) + 4 ( \sin^6 x + \cos^6 x)

= 3 [ \sin^4 x + \cos^4 x - 4 \sin x \cos x ( \sin^2 x + \cos^2 ) + 6 \sin^2 x \cos^2 x] + 6 (1 + 2 \sin x \cos x ) + 4[ (\cos^2 x +\sin^2 x)( \cos^4 x - \cos^2 x \sin^2 x+ \sin^4 x) ]

= 3 [ \sin^4 x + \cos^4 x - 4 \sin x \cos x  + 6 \sin^2 x \cos^2 x] + 6 (1 + 2 \sin x \cos x ) + 4( \cos^4 x - \cos^2 x \sin^2 x+ \sin^4 x )

= 3 \sin^4 x + 3 \cos^4 x - 12 \sin x \cos x  + 18 \sin^2 x \cos^2 x + 6  + 12 \sin x \cos x + 4 \cos^4 x - 4 \cos^2 x \sin^2 x+ 4\sin^4 x 

= 7 \sin^4 x + 7 \cos^4 x + 14 \sin^2 x \cos^2 x + 6 

= 7 (\sin^4 x +  \cos^4 x + 2 \sin^2 x \cos^2 x) + 6 

= 7 ( \sin^2 x + \cos^2 x )^2 + 6 

= 7 + 6 = 13 =  RHS. Hence proved.

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Question 20: 2 (\sin^6 x + \cos^6 x) - 3 (\sin^4 x + \cos^4 x) + 1 = 0

Answer:

LHS = 2 (\sin^6 x + \cos^6 x) - 3 (\sin^4 x + \cos^4 x) + 1

= 2 [ (\sin^3 x)^2 + (\cos^3 x)^2 ] - 3 ( \sin^4 x + \cos^4 x ) + 1

= 2 [ (\sin^2 x + \cos^2 x) ( \sin^4 x - \sin^2 x \cos^2 x + \cos^4x) ] - 3 ( \sin^4 x + \cos^4 x ) + 1

= 2 [  \sin^4 x - \sin^2 x \cos^2 x + \cos^4x ] - 3 ( \sin^4 x + \cos^4 x ) + 1

= 2 \sin^4 x - 2 \sin^2 x \cos^2 x + 2\cos^4x  - 3  \sin^4 x -3 \cos^4 x  + 1

= - \sin^4 x - \cos^4 x - 2 \sin^2 x \cos^2 x + 1

= - (\sin^2 x + \cos^2 x)^2 + 1

= -1 + 1 = 0 = RHS. Hence proved.

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Question 21: \cos^6 x - \sin^6 = \cos 2x \Big( 1 - \frac{1}{4} \sin^2 2x \Big)

Answer:

LHS = \cos^6 x - \sin^6 x

= (\cos^2 x)^3 - (\sin^2 x)^3

= (\cos^2 x - \sin^2 x) \Big[   \cos^4 x + \sin^2 x \cos^2 x + \sin^4 x \Big]

= \cos 2x  \Big[   \cos^4 x + 2\sin^2 x \cos^2 x + \sin^4 x  - \sin^2 x \cos^2 x \Big]

= \cos 2x  \Big[   (\cos^2 x + \sin^2 x)^2  - \sin^2 x \cos^2 x \Big]

= \cos 2x  \Big[   1  - \sin^2 x \cos^2 x \Big]

= \cos 2x  \Big[   1  - \frac{1}{4} ( 2\sin x \cos x)^2 \Big]

= \cos 2x  \Big[   1  - \frac{1}{4} \sin^ 2x \Big] =   RHS. Hence proved.

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Question 22: \tan \Big( \frac{\pi}{4} + x \Big) + \tan \Big( \frac{\pi}{4} - x \Big)  = 2 \sec 2x

Answer:

LHS = \tan \Big( \frac{\pi}{4} + x \Big) + \tan \Big( \frac{\pi}{4} - x \Big)

= \frac{\tan \frac{\pi}{4} + \tan x}{1 - \tan \frac{\pi}{4} \tan x} + \frac{\tan \frac{\pi}{4} 1 \tan x}{1 + \tan \frac{\pi}{4} \tan x}

= \frac{1 + \tan x}{1 -  \tan x} + \frac{1 - \tan x}{1 +  \tan x}

= \frac{(1 + \tan x)^2 - (1 - \tan x)^2}{1 - \tan^2 x}

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

= \frac{2(1 + \tan^2 x )}{1 - \tan^2 x}

= \frac{2( \cos^2 x + \sin^2 x)}{\cos^2 x - \sin^2 x}

= \frac{2}{\cos^2 x - \sin^2 x}

= \frac{2}{\cos 2x}

= 2 \sec 2x = RHS. Hence proved.

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Question 23: \cot^2 x - \tan^2 x = 4 \cot 2x \mathrm{cosec} 2x

Answer:

LHS = \cot^2 x - \tan^2 x

= \frac{\cos^2 x}{\sin^2 x} - \frac{\sin^2 x}{\cos^2 x}

= \frac{\cos^4 x - \sin^4 x}{\sin^2 x \cos^2 x}

= \frac{(\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x)}{\sin^2 x \cos^2 x}

= \frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x}

= \frac{4 \cos 2x}{(2\sin x \cos x)^2}

= \frac{4 \cos 2x}{\sin^2 2x}

= 4 \cot 2x \ \mathrm{cosec} 2x = RHS. Hence proved.

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Question 24: \cos 4x - \cos 4 \alpha = 8( \cos x - \cos \alpha) ( \cos x + \cos \alpha) (\cos x - \sin \alpha)(\cos x + \sin \alpha)

Answer:

LHS = \cos 4x - \cos 4 \alpha

= 2 \cos^2 2x - 2 \cos^2 2\alpha

= 2 ( \cos 2x + \cos 2\alpha)(\cos 2x - \cos 2 \alpha)

= 2 ( 2 \cos^2 x - 1 + 1 - 2 \sin^2 \alpha)( 2 \cos^2 x - 1 - 2 \cos^2 \alpha + 1)

= 8 (\cos^2 x - \sin^2 \alpha) ( \cos^2 x - \cos^2 \alpha)

= 8 ( \cos x - \cos \alpha)(\cos x + \cos \alpha) ( \cos x + \cos\alpha)(\cos x - \cos \alpha)

= RHS. Hence proved.

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Question 25: \sin 3x + \sin 2x - \sin x = 4 \sin x \cos \frac{x}{2} \cos \frac{3x}{2}

Answer:

LHS = \sin 3x + \sin 2x - \sin x

= ( \sin 3x - \sin x) + \sin 2x

= 2 \cos 2x \sin x + 2 \sin x \cos x

= 2 \sin x ( \cos 2x + \cos x)

= 2 \sin x \Big( 2 \cos \frac{3x}{2} \cos \frac{x}{2} \Big)

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

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Question 26: Prove that: \tan 82 \frac{1}{2} = (\sqrt{3}+\sqrt{2})(\sqrt{2}+1) = \sqrt{2}+ \sqrt{3}+ \sqrt{4}+ \sqrt{6}

Answer:

We know \tan \frac{x}{2} = \frac{\sin x}{1+\cos x}

\Rightarrow \tan 7 \frac{1}{2} ^o = \frac{\sin 15^o}{1+\cos 15^o}

= \frac{\sin (45^o-30^o)}{1+\cos (45^o-30^o)}

= \frac{\sin 45^o \cos 30^o - \sin 30^o \cos 45^o}{1 + 45^o \cos 30^o + \sin 45^o \sin 30^o}

= \frac{\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} }

= \frac{\frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2 \sqrt{2}} }{1 + \frac{\sqrt{3}}{2\sqrt{2}} \times \frac{1}{2 \sqrt{2}} }

= \frac{\sqrt{3}-1}{2\sqrt{2} + \sqrt{3} + 1}

\therefore \tan 7 \frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2} + \sqrt{3} + 1}

\therefore \cot 7 \frac{1}{2} = \frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3}-1}

\Rightarrow \cot (90 - 82 \frac{1}{2} ) = \frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3}-1}

\Rightarrow \tan 82 \frac{1}{2} = \frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3}-1}

= \frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}

= \frac{2\sqrt{6} + 3 + \sqrt{3}+ 2\sqrt{2}+\sqrt{3}+1}{2}

= \frac{2\sqrt{6} + 4 + 2\sqrt{3}+ 2\sqrt{2}}{2}

= \sqrt{6} + 2 + \sqrt{3}+ \sqrt{2}

= \sqrt{2} + \sqrt{3} + \sqrt{4}+ \sqrt{5} = RHS. Hence proved.

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Question 27: Prove that \cot \frac{\pi}{8} = \sqrt{2}+1

Answer:

LHS = \cot \frac{\pi}{8} = \cot 22 \frac{1}{2}

We know, \cot 2x = \frac{\cot^2 x - 1}{2 \cot x}

\Rightarrow \cot 45^o = \frac{\cot^2 22\frac{1}{2} - 1}{2 \cot 22\frac{1}{2}}

\Rightarrow 2 \cot 22 \frac{1}{2} = \cot^2 22 \frac{1}{2} - 1

Let \cot 22 \frac{1}{2} = a

\Rightarrow a^2 - 2a -1 = 0

\Rightarrow (a-1)^2 = 2

\Rightarrow a- 1 = \pm \sqrt{2}

\Rightarrow a = 1 + \pm \sqrt{2}

Since \cot 22\frac{1}{2}   is in I quadrant, it is positive

\therefore \cot 22\frac{1}{2}  = 1 + \sqrt{2} = RHS. Hence proved.

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Question 28: i) If \cos x = - \frac{3}{5} and x lies in III quadrant, find the value of \cos \frac{x}{2} , \sin \frac{x}{2} and \sin 2x

ii) If \cos x = - \frac{3}{5} and x lies in II quadrant, find the value of \sin 2x and \sin \frac{x}{2} .

Answer:

i) Given \cos x = - \frac{3}{5}

We know \cos 2x = \cos^2 x - \sin^2 x

\Rightarrow \cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}

\Rightarrow - \frac{3}{5} = 2 \cos^2 \frac{x}{2} - 1

\Rightarrow \cos^2 \frac{x}{2} = \frac{1}{2} \Big[ - \frac{3}{5} + 1 \Big]

\Rightarrow \cos^2 \frac{x}{2} = \frac{1}{5}

\Rightarrow \cos \frac{x}{2} = \pm \frac{1}{\sqrt{5}}

Given that \frac{x}{2} lies in II Quadrant, \cos \frac{x}{2} is negative.

\Rightarrow \cos \frac{x}{2} = - \frac{1}{\sqrt{5}}

Similarly,

\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}

\Rightarrow - \frac{3}{5} = 1 - 2 \sin^2 \frac{x}{2}

\Rightarrow 2 \sin^2 \frac{x}{2} = 1 + \frac{3}{5}

\Rightarrow \sin^2 \frac{x}{2} = \frac{4}{5}

\Rightarrow \sin \frac{x}{2} = \pm \frac{2}{\sqrt{5}}

Given that \frac{x}{2} lies in II Quadrant, \sin \frac{x}{2} is positive.

\Rightarrow \sin \frac{x}{2} = \frac{2}{\sqrt{5}}

Now \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \big( - \frac{3}{5} \big)^2 } = \pm \frac{4}{5}

Given that x lies in III Quadrant, \sin x   is negative.

\therefore \sin x = - \frac{4}{5}

Also \sin 2x = 2 \sin x \cos x = 2 \big( - \frac{4}{5} \big) \big( - \frac{3}{5} \big) = \frac{24}{25}

ii) Given \cos x = - \frac{3}{5} and x lies in II Quadrant

\therefore \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \big( -\frac{3}{5} \big)^2} = \pm \frac{4}{5}

Since x lies in II Quadrant, \sin x is positive

\therefore \sin x =  \frac{4}{5}

We know \sin 2x = 2 \sin x \cos x

\Rightarrow \sin 2x = 2 \big( \frac{4}{5} \big) \big( - \frac{3}{5} \big)  = - \frac{24}{25}

We know \cos x = 1 - 2 \sin^2 \frac{x}{2}

\Rightarrow \sin^2 \frac{x}{2} = \frac{1}{2} ( 1 - \cos x) = \frac{1}{2} \Big( 1- \big( - \frac{3}{5} \big) \Big) = \frac{4}{5}

\Rightarrow \sin \frac{x}{2}= \pm \frac{2}{\sqrt{5}}

Since \frac{x}{2} lies in I Quadrant, \sin \frac{x}{2} is positive

\therefore \sin \frac{x}{2} = \frac{2}{\sqrt{5}}

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Question 29: If \sin x =  \frac{\sqrt{5}}{3} and x lies in II quadrant, find the value of \cos \frac{x}{2} ,     \sin \frac{x}{2} and \tan \frac{x}{2}

Answer:

Given \sin x = \frac{\sqrt{5}}{3}

\Rightarrow \cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 -\frac{5}{9}} = \pm \frac{2}{3}

Since x lies in II Quadrant , \cos x is positive

\therefore \cos x = - \frac{2}{3}

We know \cos x = 1 - 2 \sin^2 \frac{x}{2}

\sin^2 \frac{x}{2} = \frac{1}{2} \Big[ 1 - \big( - \frac{2}{3} \big) \Big] = \frac{5}{6}

\sin \frac{x}{2} = \pm \sqrt{\frac{5}{6}}

Since \frac{x}{2} lies in I Quadrant , \sin \frac{x}{2} is positive

\therefore \sin \frac{x}{2} = \sqrt{\frac{5}{6}}

We know \cos x = 2 \cos^2 \frac{x}{2} - 1

\Rightarrow  \cos^2 \frac{x}{2} = \frac{1}{2} \Big[ \big( - \frac{2}{3} \big) + 1 \Big] = \frac{1}{2} \Big[ \frac{1}{3} \Big] = \frac{1}{6}

\therefore \cos \frac{x}{2} = \pm \frac{1}{\sqrt{6}}

Since \frac{x}{2} lies in I Quadrant , \cos \frac{x}{2} is positive

\therefore \cos \frac{x}{2} = \frac{1}{\sqrt{6}}

\therefore \tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\sqrt{\frac{5}{6}} }{\frac{1}{\sqrt{6}}} = \sqrt{5}

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Question 30:

i) if 0 \leq x \leq \pi   and x lie in II quadrant such that \sin x = \frac{1}{4} , find the value of \cos \frac{x}{2} , \sin \frac{x}{2} and \tan \frac{x}{2}

ii) If \cos x = \frac{4}{5} and x is acute, find \tan 2x

iii) If \sin x = \frac{4}{5} and 0 < x < \frac{\pi}{2} , find the value of  \sin 4x

Answer:

i) Given \sin x = \frac{1}{4}

\therefore \cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \frac{1}{16}} = \pm \frac{\sqrt{15}}{4}

Since x lie in II Quadrant, \cos x  is negative

\therefore \cos x = - \frac{\sqrt{15}}{4}

Now we know, \cos x = 2 \cos^2 \frac{x}{2} - 1

\Rightarrow \cos^2 \frac{x}{2} = \frac{1}{2} \Big[ - \frac{\sqrt{15}}{4} + 1 \Big] = \frac{4 - \sqrt{15}}{8}

\Rightarrow \cos \frac{x}{2} = \pm \sqrt{ \frac{4 - \sqrt{15}}{8}}

Since x lies in II Quadrant, \frac{x}{2}   will lie in I Quadrant. Hence \cos \frac{x}{2} is positive.

\Rightarrow \cos \frac{x}{2} =  \sqrt{ \frac{4 - \sqrt{15}}{8}}

Now \cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}

\Rightarrow - \frac{\sqrt{15}}{4} = \Big( \sqrt{ \frac{4 - \sqrt{15}}{8}} \Big)^2 -  \sin^2 \frac{x}{2}

\sin^2 \frac{x}{2} = \frac{4 - \sqrt{15}}{8} + \frac{\sqrt{15}}{4}

\sin^2 \frac{x}{2} = \frac{4 + \sqrt{15}}{8}

\Rightarrow \sin \frac{x}{2} = \pm \sqrt{ \frac{4 + \sqrt{15}}{8}}

Since x lies in II Quadrant, \frac{x}{2}   will lie in I Quadrant. Hence \sin \frac{x}{2} is positive.

\Rightarrow \sin \frac{x}{2} =  \sqrt{ \frac{4 + \sqrt{15}}{8}}

\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\sqrt{ \frac{4 + \sqrt{15}}{8}}}{\sqrt{ \frac{4 - \sqrt{15}}{8}}} = \sqrt{ \frac{4 + \sqrt{15}}{4 - \sqrt{15}} } = \sqrt{ \frac{4 + \sqrt{15}}{4 - \sqrt{15}} \times \frac{4 + \sqrt{15}}{4 + \sqrt{15}} } = 4 + \sqrt{15}

ii) Given \cos x = \frac{4}{5}

\tan x = \frac{\sin x }{\cos x} = \frac{\sqrt{1 - \cos^2 x}}{\cos x} = \frac{\sqrt{1 - \big( \frac{4}{5} \big)^2}}{\frac{4}{5}} = \frac{3}{4}

\tan 2x = \frac{2 \tan x}{1 - \tan^2 x} = \frac{2 \times \frac{3}{4}}{1 - \big( \frac{3}{4} \big)^2} = \frac{24}{7}

iii) \sin x = \frac{4}{5} and 0 < x < \frac{\pi}{2} 

\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}

\therefore \sin 2x = 2 \sin x \cos x = 2 \times \frac{4}{5} \times \frac{3}{5} = \frac{24}{25}

\cos 2x = \sqrt{1 - \sin^2 2x} = \sqrt{1 - \big( \frac{24}{25} \big)^2} = \pm \frac{7}{25}

Since x likes in I Quadrant, 2x lies in II Quadrant \therefore \cos 2x is negative

\Rightarrow \cos 2x = - \frac{7}{25}

Now \sin 4x = 2 \sin 2x \cos 2x = 2 \times \frac{24}{7} \times \big( - \frac{7}{25} \big) = - \frac{336}{625}

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Question 31: If \tan x = \frac{b}{a} , find the value of \sqrt{\frac{a+b}{a-b}} + \sqrt{\frac{a-b}{a+b}}

Answer:

\sqrt{\frac{a+b}{a-b}} + \sqrt{\frac{a-b}{a+b}}

= \frac{a+b + a - b}{\sqrt{a^2 - b^2}}

= \frac{2a}{\sqrt{a^2 - b^2}}

= \frac{2}{\sqrt{1 - \big(\frac{b}{a} \big)^2}}

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

= \frac{2 \cos x}{\sqrt{\cos^2 x - \sin^2 x} }

= \frac{2 \cos x}{\sqrt{\cos 2x}}

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Question 32: If \tan A = \frac{1}{y} and If \tan B = \frac{1}{3} , show that \cos 2A = \sin 4B

Answer:

Given \tan A = \frac{1}{y} and If \tan B = \frac{1}{3}

\cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{1 - \big( \frac{1}{y} \big)^2}{1 + \big( \frac{1}{y} \big)^2} = \frac{49-1}{49+1} = \frac{48}{50} = \frac{24}{25}

\sin 4B = 2 \sin 2B \cos 2B

= 2 \Big(  \frac{2\tan B}{1 + \tan^2 B} \Big) \Big(  \frac{1 - \tan^2 B}{1 + \tan^2 B} \Big)

= 4 \Big(  \frac{\frac{1}{3}}{1 + \big( \frac{1}{3}\big)^2 B} \Big) \Big(  \frac{1 - \big( \frac{1}{3}\big)^2}{1 + \big( \frac{1}{3}\big)^2} \Big)   = \frac{4 \times \frac{1}{3} \times \frac{8}{9}}{\frac{10}{9} \times \frac{10}{9} }   = \frac{4 \times 3 \times 8}{10 \times 10} = \frac{24}{25}

Hence \cos 2A = \sin 4B

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Question 33: Prove that \cos 7^o \cos 14^o \cos 28^o \cos 56^o = \frac{\sin 68^o}{16 \cos 83^o}

Answer:

LHS = \cos 7^o \cos 14^o \cos 28^o \cos 56^o

= \frac{2 \sin 7^o \cos 7^o \cos 14^o \cos 28^o \cos 56^o}{2 \sin 7^o }

= \frac{2 \sin 14^o  \cos 14^o \cos 28^o \cos 56^o}{4 \sin 7^o }

= \frac{2 \sin 28^o   \cos 28^o \cos 56^o}{8 \sin 7^o }

= \frac{2 \sin 56^o    \cos 56^o}{16 \sin 7^o }

= \frac{\sin 112^o}{16 \sin 7^o }

= \frac{\sin (180^o - 68^o)}{16 \sin (90^o - 83^o) }

= \frac{\sin 68^o}{16 \cos 83^o } = RHS. Hence proved.

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Question 34: Prove that \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15} = \frac{1}{16}

Answer:

LHS = \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15}

= \cos 24^o \cos 48^o \cos 96^o \cos 192^o

= \frac{2 \sin 24^o \cos 24^o \cos 48^o \cos 96^o \cos 192^o}{2 \sin 24^o}

= \frac{2 \sin 48^o \cos 48^o \cos 96^o \cos 192^o}{4 \sin 24^o}

= \frac{2 \sin 96^o  \cos 96^o \cos 192^o}{8 \sin 24^o}

= \frac{2 \sin 192^o \cos 192^o}{16 \sin 24^o}

= \frac{\sin 384^o}{16 \sin 24^o}

= \frac{\sin (360^o+24^o)}{16 \sin 24^o}

= \frac{\sin 24^o}{16 \sin 24^o}

= \frac{1}{16} = RHS. Hence proved.

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Question 35: Prove that \cos \frac{\pi}{5} \cos \frac{2\pi}{5} \cos \frac{4\pi}{5} \cos \frac{8\pi}{5} = - \frac{1}{16}

Answer:

LHS = \cos \frac{\pi}{5} \cos \frac{2\pi}{5} \cos \frac{4\pi}{5} \cos \frac{8\pi}{5}

= \cos 12^o \cos 24^o \cos 48^o \cos 96^o

= \frac{2 \sin 12^o \cos 12^o \cos 24^o \cos 48^o \cos 96^o}{2 \sin 12^o}

= \frac{2 \sin 24^o \cos 24^o \cos 48^o \cos 96^o}{4 \sin 12^o}

= \frac{2 \sin 48^o  \cos 48^o \cos 96^o}{8 \sin 12^o}

= \frac{2 \sin 96^o \cos 96^o}{16 \sin 12^o}

= \frac{\sin 192^o}{16 \sin 12^o}

= \frac{\sin (180^o+12^o)}{16 \sin 12^o}

= \frac{-\sin 12^o}{16 \sin 12^o}

= - \frac{1}{16} = RHS. Hence proved.

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Question 36: Prove that \cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos \frac{4\pi}{65} \cos \frac{8\pi}{65} \cos \frac{16\pi}{65} \cos \frac{32\pi}{65} = \frac{1}{64}

Answer:

LHS = \cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos \frac{4\pi}{65} \cos \frac{8\pi}{65} \cos \frac{16\pi}{65} \cos \frac{32\pi}{65}

= \frac{2 \sin \frac{\pi}{65} \cos \frac{\pi}{65} \cos  \frac{2\pi}{65}  \cos  \frac{4\pi}{65}  \cos  \frac{8\pi}{65}  \cos  \frac{16\pi}{65}}{2 \sin \frac{\pi}{65}}

= \frac{2 \sin \frac{2\pi}{65}  \cos  \frac{2\pi}{65}  \cos  \frac{4\pi}{65}  \cos  \frac{8\pi}{65}  \cos  \frac{16\pi}{65}  \cos  \frac{32\pi}{65}   }{4 \sin \frac{\pi}{65}}

= \frac{2 \sin \frac{4\pi}{65}    \cos  \frac{4\pi}{65}  \cos  \frac{8\pi}{65}  \cos  \frac{16\pi}{65} \cos  \frac{32\pi}{65}}{8 \sin \frac{\pi}{65}}

= \frac{2 \sin \frac{8\pi}{65}     \cos  \frac{8\pi}{65}  \cos  \frac{16\pi}{65} \cos  \frac{32\pi}{65} }{16 \sin \frac{\pi}{65}}

= \frac{2 \sin \frac{16\pi}{65}   \cos  \frac{16\pi}{65} \cos  \frac{32\pi}{65} }{32 \sin \frac{\pi}{65}}

= \frac{2 \sin \frac{32\pi}{65}  \cos  \frac{32\pi}{65} }{64 \sin \frac{\pi}{65}}

= \frac{ \sin \frac{64\pi}{65}   }{64 \sin \frac{\pi}{65}}

= \frac{ \sin \big( \pi - \frac{\pi}{65} \big)   }{64 \sin \frac{\pi}{65}}

= \frac{ \sin \frac{\pi}{65}   }{64 \sin \frac{\pi}{65}}

= \frac{1}{64} = RHS. Hence proved.

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Question 37: If 2 \tan \alpha = 3 \tan \beta , then prove that \tan (\alpha - \beta) = \frac{\sin 2 \beta}{5- \cos 2\beta}

Answer:

Given 2 \tan \alpha = 3 \tan \beta \Rightarrow \tan \alpha = \frac{3}{2} \tan \beta

LHS = \tan (\alpha - \beta)

= \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}

= \frac{\frac{3}{2} \tan \beta - \tan \beta}{1 + \frac{3}{2} \tan \beta \tan \beta}

= \frac{\tan \beta}{2 + 3 \tan^2 \beta}

= \frac{\frac{\sin \beta}{\cos \beta}}{2 + 3 \big( \frac{\sin \beta}{\cos \beta} \big)^2}

= \frac{\sin \beta \cos \beta}{2 \cos^2 \beta + 3 \sin^2 \beta}

= \frac{\sin \beta \cos \beta}{2 + \sin^2 \beta}

= \frac{2\sin \beta \cos \beta}{4 + 2 \sin^2 \beta}

= \frac{\sin 2\beta}{4 + 2 (1- \cos^2 \beta)}

= \frac{\sin 2\beta}{6 - 2 \cos^2 \beta} = RHS. Hence proved.

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Question 38: If  \sin \alpha + \sin \beta = a and  \cos \alpha + \cos \beta = b , prove that:

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

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

Answer:

i) Given \sin \alpha + \sin \beta = a

\Rightarrow 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}  = a    … … … … … i)

Given   \cos \alpha + \cos \beta = b

\Rightarrow 2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} = b     … … … … … ii)

Dividing i) by ii) we get

\tan \frac{\alpha + \beta}{2} = \frac{a}{b}

We know \sin (\alpha + \beta) = \frac{2 \tan \big( \frac{\alpha + \beta}{2} \big)}{1+ \tan^2 \big( \frac{\alpha + \beta}{2} \big)} = \frac{2 \frac{a}{b}}{1+\frac{a^2}{b^2} } = \frac{2ab}{a^2+b^2}

ii) Squaring and adding

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

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

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

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

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

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

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Question 39: If  2\tan \frac{\alpha}{2} = \tan \frac{\beta}{2} , prove that \cos \alpha = \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}

Answer:

Given 2\tan \frac{\alpha}{2} = \tan \frac{\beta}{2} ,

RHS = \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}

We know \cos \beta = \frac{1 - \tan^2 \frac{\beta}{2} }{1 + \tan^2 \frac{\beta}{2} }

= \frac{3 + 5 \big( \frac{1 - \tan^2 \frac{\beta}{2} }{1 + \tan^2 \frac{\beta}{2} } \big) }{5 + 3 \big(\frac{1 - \tan^2 \frac{\beta}{2} }{1 + \tan^2 \frac{\beta}{2} } \big)}

= \frac{3 + 3 \tan^2 \frac{\beta}{2}+ 5 - 5 \tan^2 \frac{\beta}{2}}{5 + 5\tan^2 \frac{\beta}{2} + 3 - 3 \tan^2 \frac{\beta}{2}}

= \frac{8 - 2 \tan^2 \frac{\beta}{2}}{8 + 2 \tan^2 \frac{\beta}{2}}

Since \tan \frac{\beta}{2} = 2 \tan \frac{\alpha}{2}

= \frac{8 - 2 \big( 2 \tan \frac{\alpha}{2} \big)^2 }{8 + 2 \big( 2 \tan \frac{\alpha}{2} \big)^2 }

= \frac{8 - 8 \tan^2 \frac{\alpha}{2}}{8 + 8 \tan^2 \frac{\alpha}{2}}

= \frac{1 -  \tan^2 \frac{\alpha}{2}}{1 +  \tan^2 \frac{\alpha}{2}}

= \cos \alpha = LHS. Hence proved.

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Question 40: If \cos x = \frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta} prove that \tan \frac{x}{2} = \pm \tan \frac{\alpha}{2} \tan \frac{\beta}{2}

Answer:

Given \cos x = \frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}

\Rightarrow  \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} = \frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}

Applying Componendo and Dividendo

\frac{1 - \tan^2 \frac{x}{2}+ 1 + \tan^2 \frac{x}{2}}{1 - \tan^2 \frac{x}{2} -1 - \tan^2 \frac{x}{2}} = \frac{\cos \alpha + \cos \beta + 1 + \cos \alpha \cos \beta}{\cos \alpha + \cos \beta  - 1 - \cos \alpha \cos \beta}

\Rightarrow \frac{2}{-2 \tan^2 \frac{x}{2}} = \frac{(1 + \cos \alpha)(1 + \cos \beta)}{- (1 - \cos \alpha)(1 - \cos \beta)}

\Rightarrow \tan^2 \frac{x}{2} = \frac{(1 - \cos \alpha)(1 - \cos \beta)}{ (1 + \cos \alpha)(1 + \cos \beta)}

\Rightarrow \tan^2 \frac{x}{2} = \frac{2 \sin^2 \frac{\alpha}{2} \sin^2 \frac{\beta}{2} }{ 2 \cos^2 \frac{\alpha}{2} \cos^2 \frac{\beta}{2} }

\Rightarrow \tan^2 \frac{x}{2} = \tan^2 \frac{\alpha}{2} \tan^2 \frac{\beta}{2}

\Rightarrow \tan \frac{x}{2} = \pm \tan \frac{\alpha}{2} \tan \frac{\beta}{2}

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Question 41: If \sec (x + \alpha) + \sec ( x - \alpha) = 2 \sec x , prove that \cos x = \pm \sqrt{2} \cos \frac{\alpha}{2}

Answer:

Given \sec (x + \alpha) + \sec ( x - \alpha) = 2 \sec x , prove that \cos x = \pm \sqrt{2} \cos \frac{\alpha}{2}

\Rightarrow \frac{1}{\cos (x + \alpha)} + \frac{1}{\cos (x - \alpha)} = \frac{2}{\cos x}

\Rightarrow \frac{\cos (x + \alpha) + \cos (x - \alpha)}{\cos (x + \alpha)\cos (x - \alpha)} = \frac{2}{\cos x}

\Rightarrow \frac{2 \cos x \cos \alpha}{\cos^2 x \cos^2 \alpha - \sin^2 x \sin^2 \alpha } = \frac{2}{\cos x}

\Rightarrow \frac{\cos^2 x \cos \alpha}{\cos^2 x \cos^2 \alpha - \sin^2 x \sin^2 \alpha } = 1

\Rightarrow \cos^2 x \cos \alpha = \cos^2 x \cos^2 \alpha - \sin^2 x \sin^2 \alpha

\Rightarrow \cos^2 x \cos \alpha = \cos^2 x \cos^2 \alpha - (1 - \cos^2 x) \sin^2 \alpha

\Rightarrow \cos^2 x \cos \alpha = \cos^2 x \cos^2 \alpha - \sin^2 \alpha + \cos^2 x  \sin^2 \alpha

\Rightarrow \cos^2 x \cos \alpha = \cos^2 x ( \cos^2 \alpha + \sin^2 \alpha) -  \sin^2 \alpha

\Rightarrow \cos^2 x \cos \alpha = \cos^2 x  -  \sin^2 \alpha

\Rightarrow \cos^2 x \cos \alpha - \cos^2 x  = -  \sin^2 \alpha

\Rightarrow \cos^2 x ( \cos \alpha -1)  = -  \sin^2 \alpha

\Rightarrow \cos^2 x ( 1 - \cos \alpha )  =  \sin^2 \alpha

\Rightarrow \cos^2 x = \frac{\sin^2 \alpha}{ 1 - \cos \alpha }

\Rightarrow \cos^2 x= \frac{4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2} }{ 2 \sin^2 \frac{\alpha}{2} }

\Rightarrow \cos^2 x = 2 \cos^2 \frac{\alpha}{2}

\Rightarrow \cos x = \pm \sqrt{2} \cos \frac{\alpha}{2} . Hence proved.

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Question 42: \cos \alpha + \cos \beta = \frac{1}{3} and \sin \alpha + \sin \beta = \frac{1}{4} , prove that \cos \frac{\alpha - \beta}{2} = \pm \frac{5}{24}

Answer:

Given \cos \alpha + \cos \beta = \frac{1}{3} and \sin \alpha + \sin \beta = \frac{1}{4}

Squaring and adding the two

(\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2 = \frac{1}{9} + \frac{1}{16}

\Rightarrow \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta  + \sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta = \frac{1}{9} + \frac{1}{16}

\Rightarrow 2 + 2 \cos \alpha \cos \beta + 2 \sin \alpha \sin \beta = \frac{25}{144}

\Rightarrow 2 + 2 \cos( \alpha - \beta) = \frac{25}{144}

\Rightarrow \cos( \alpha - \beta) = - \frac{263}{288}

Since \cos^2 \big( \frac{\alpha - \beta}{2} \big) = \frac{1+ \cos ( \alpha - \beta)}{2} = \frac{1- \frac{263}{288}}{2} = \frac{25}{576}

\Rightarrow \cos \big( \frac{\alpha - \beta}{2} \big) = \pm \frac{5}{24}

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Question 43: If \sin \alpha = \frac{4}{5} and \cos \beta = \frac{5}{13} , prove that \cos \frac{\alpha - \beta}{2} = \frac{8}{\sqrt{65}}

Answer:

Given \sin \alpha = \frac{4}{5}

\Rightarrow \cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}

Also given \cos \beta = \frac{5}{13}

\Rightarrow \sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - \frac{25}{169}} = \frac{12}{13}

\cos( \alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta

\Rightarrow \cos ( \alpha - \beta) = \frac{3}{5} \times \frac{5}{13} +  \frac{4}{5} \times \frac{12}{13}   = \frac{63}{65}

Therefore \cos \frac{ \alpha - \beta}{2} = \sqrt{\frac{1+ \cos ( \alpha - \beta)}{2} } = \sqrt{\frac{1+\frac{63}{65}}{2}} = \pm \frac{8}{\sqrt{65}}

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Question 44: If a \cos 2x + b \sin 2x = c has \alpha and \beta as its roots, then prove that:

i) \tan \alpha + \tan \beta = \frac{2b}{a+c}      

ii) \tan \alpha \tan \beta = \frac{c-a}{c+a}      

iii) \tan ( \alpha + \beta) = \frac{b}{a}

Answer:

i) Given a \cos 2x + b \sin 2x = c      … … … … … i)

\Rightarrow a \Big( \frac{1 - \tan^2 x}{1 + \tan^2 x} \Big) + b \Big( \frac{2 \tan x}{1 + \tan^2 x} \Big) = c

\Rightarrow a ( 1 - \tan^2 x ) + 2 b \tan x = c ( 1 + \tan^2 x)

\Rightarrow a  - a \tan^2 x  + 2 b \tan x = c  + c \tan^2 x

\Rightarrow (c+a) \tan^2 x - 2 b \tan x + ( c-a) = 0      … … … … … ii)

if \alpha and \beta are the roots of i) , then \tan \alpha and \tan \beta are the roots of ii)

\therefore \tan \alpha + \tan \beta = \frac{-(-2b)}{a+c} = \frac{2b}{a+c}

ii) Similarly \tan \alpha  \tan \beta = \frac{c-a}{c+a}

iii) \tan ( \alpha - \beta ) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha  \tan \beta} = \frac{\frac{2b}{a+c}}{1 - \frac{c-a}{c+a} } = \frac{2b}{a+c-c+a} = \frac{b}{a}

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Question 45: If \cos \alpha + \cos \beta = 0 = \sin \alpha + \sin \beta , then prove that \cos 2\alpha + \cos 2\beta= -2 \cos (\alpha + \beta)

Answer:

Given \cos \alpha + \cos \beta = 0 = \sin \alpha + \sin \beta

Squaring both sides

(\cos \alpha + \cos \beta)^2 = (\sin \alpha + \sin \beta)^2

\cos^2 \alpha + \cos^2 \beta +2 \cos \alpha  \cos \beta = \sin^2 \alpha + \sin^2 \beta +2 \sin \alpha  \sin \beta

\cos^2 \alpha - \sin^2 \alpha + \cos^2 \beta - \sin^2 \beta = -2 ( - \sin \alpha  \sin \beta + \cos \alpha  \cos \beta )

2 \cos^2 \alpha - 1 + 2 \cos^2 \beta - 1 = -2 \cos ( \alpha + \beta)

\cos 2\alpha + \cos 2 \beta = - 2 \cos ( \alpha + \beta)

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