Prove the following identities (1-16) :

Question 1: \sec^4 x - \sec^2 x = \tan^4 x + \tan^2 x

Answer:

LHS = \sec^4 x - \sec^2 x

= \sec^2 x ( \sec^2 x - 1)

= (1+ \tan^2 x)(\tan^2 x)

= \tan^4 x + \tan^2 x = RHS. Hence proved.

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Question 2: \sin^6 x + \cos^6 x = 1 - 3 \sin^2 x \cos^2 x

Answer:

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

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

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

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

=  1 - 3 \sin^2 x \cos^2 x = RHS. Hence proved.

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Question 3: (\mathrm{cosec} x - \sin x)(\sec x - \cos x)(\tan x + \cot x) = 1

Answer:

LHS = (\mathrm{cosec} x - \sin x)(\sec x - \cos x)(\tan x + \cot x)

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

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

= \frac{\cos^2 x}{\sin x} . \frac{\sin^2 x}{\cos x} . \frac{1}{\sin x \cos x}

= 1 = RHS. Hence proved.

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Question 4: \mathrm{cosec} x (\sec x - 1) - \cot x ( 1- \cos x) = \tan x - \sin x

Answer:

LHS = \mathrm{cosec} x (\sec x - 1) - \cot x ( 1- \cos x)

= \frac{1}{\sin x} \Big( \frac{1}{\cos x} -1 \Big) - \frac{\cos x }{\sin x} (1- \cos x)

= \frac{1-\cos x}{\sin x \cos x} - \frac{(1- \cos x) \cos x}{\sin x}

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

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

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

= \frac{\sin x}{\cos x} (1-\cos x)

= \tan x - \sin x = RHS. Hence proved.

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Question 5: \frac{1 - \sin x \cos x}{\cos x (\sec x - \mathrm{cosec} x)} . \frac{\sin^2 x - \cos^2 x}{\sin^3 x + \cos^3 x} = \sin x

Answer:

LHS = \frac{1 - \sin x \cos x}{\cos x (\sec x - \mathrm{cosec} x)} . \frac{\sin^2 x - \cos^2 x}{\sin^3 x + \cos^3 x}

= \frac{1 - \sin x \cos x}{\cos x (\frac{1}{\cos x}- \frac{1}{\sin x})} . \frac{\sin^2 x - \cos^2 x}{\sin^3 x + \cos^3 x}

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

= \frac{ \sin x ( 1 - \sin x \cos x)}{(\sin x - \cos x)} . \frac{(\sin x - \cos x)}{(1 - \sin x \cos x)}

= \sin x = RHS. Hence proved.

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Question 6: \frac{\tan x}{1- \cot x} + \frac{\cot x}{1- \tan x} = (\sec x \ \mathrm{cosec} x + 1)

Answer:

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

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

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

= \frac{\sin^3 x - \cos^3 x}{\cos x \sin x ( \sin x - \cos x)}

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

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

= 1 + \sec x \ \mathrm{cosec} x=  RHS. Hence proved.

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Question 7: \frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} + \frac{\sin^3 x - \cos^3 x}{\sin x - \cos x} = 2

Answer:

LHS = \frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} + \frac{\sin^3 x - \cos^3 x}{\sin x - \cos x}

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

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

= 2 = RHS. Hence proved.

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Question 8: (\sec x \ \sec y + \tan x \ \tan y )^2 - (\sec x \ \tan y + \tan x \ \sec y)^2 = 1

Answer:

LHS = (\sec x \ \sec y + \tan x \ \tan y )^2 - (\sec x \ \tan y + \tan x \ \sec y)^2

=[ \sec^2 x \sec^2 y + \tan^2 x \tan^2 y + 2 \sec x \sec y \tan x \tan y ] - [ \sec^2 x \tan^2 y + \tan^2 x \sec^2 y + 2 \sec x \tan x \sec y \tan y ]

= \sec^2 x \sec^2 y + \tan^2 x \tan^2 y - \sec^2 x \tan^2 y - \tan^2 x \sec^2 y

= \sec^2 x (sec^2 y - \tan^2 y)+ \tan^2 x ( \tan^2 y - \sec^2 y)

= sec^2 x + \tan^2 x = 1 = RHS. Hence proved.

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Question 9: \frac{\cos x}{1 - \sin x} = \frac{1 + \cos x + \sin x}{1 + \cos x - \sin x}

Answer:

RHS = \frac{1 + \cos x + \sin x}{1 + \cos x - \sin x}

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

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

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

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

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

= \frac{1 + \sin x}{\cos x} \times \frac{1-\sin x}{1- \sin x}

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

= \frac{\cos x}{1- \sin x} = LHS. Hence proved.

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Question 10: \frac{\tan^3 x}{1 + \tan^2 x} + \frac{\cot^3 x}{1 + \cot^2 x} = \frac{1 - 2 \sin^2 x \cos^2 x}{\sin x \ \cos x}

Answer:

LHS = \frac{\tan^3 x}{1 + \tan^2 x} + \frac{\cot^3 x}{1 + \cot^2 x}

= \frac{\sin^3 x}{\cos^3 x (1 + \frac{\sin^2 x}{\cos^2 x})} + \frac{\cos^3 x}{\sin^3 x (1 + \frac{\cos^2 x}{\sin^2 x})}

= \frac{ \sin^3 x}{\cos x( \sin x + \cos x)} + \frac{\cos^3 x}{\sin x ( \sin x + \cos x)}

= \frac{\sin^3 x }{\cos x} + \frac{\cos^3 x}{\sin x}

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

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

= \frac{1 - 2 \sin^2 x \cos^2 x}{\sin x \cos x} =  RHS. Hence Proved.

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Question 11: 1 - \frac{\sin^2 x}{1 + \cot x} - \frac{\cos^2 x}{1 + \tan x} = \sin x \cos x  

Answer:

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

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

= 1 - \frac{\sin^3 x}{(\sin x + \cos x)} - \frac{\cos^3 x}{(\sin x + \cos x)}

= \frac{\sin x + \cos x - \sin^3 x - \cos^3 x}{(\sin x + \cos x)}

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

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

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

= \sin x \cos x =  RHS. Hence Proved.

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Question 12: \Big(  \frac{1}{\sec^2 x - \cos^2 x} + \frac{1}{ \mathrm{cosec}^2 x - \sin^2 x}    \Big) \sin^2 x \cos^2 = \frac{1- \sin^2 x \cos^2 x }{2 + \sin^2 x \cos^2 x}  

Answer:

LHS = \Big(  \frac{1}{\sec^2 x - \cos^2 x} + \frac{1}{ \mathrm{cosec}^2 x - \sin^2 x}    \Big) \sin^2 x \cos^2

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

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

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

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

= \Big( \frac{ \cos^4 x ( 1 + \sin^2 x) + \sin^4 x ( 1 + \cos^2 x)}{\sin^2 x(1+ \cos^2 x) \cos^2 x(1+ \sin^2 x) } \Big) \sin^2 x \cos^2

= \Big( \frac{ \cos^4 x ( 1 + \sin^2 x) + \sin^4 x ( 1 + \cos^2 x)}{(1+ \cos^2 x)(1+ \sin^2 x) } \Big)

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

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

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

= \Big( \frac{ 1 -  \sin^2 x \cos^2 x  }{2 + \cos^2 \sin^2 x } \Big) =  RHS. Hence Proved.

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Question 13: (1 + \tan \alpha \tan \beta)^2 + (\tan \alpha - \tan \beta)^2 = \sec^2 \alpha \sec^2 \beta

Answer:

LHS = (1 + \tan \alpha \tan \beta)^2 + (\tan \alpha - \tan \beta)^2

= 1 + \tan^2 \alpha \tan^2 \beta + 2\tan \alpha \tan \beta + \tan^2 \alpha + \tan^2 \beta - 2\tan \alpha \tan \beta

= 1 + \tan^2 \alpha \tan^2 \beta + \tan^2 \alpha + \tan^2 \beta

= 1 + \tan^2 \beta + \tan^2 \alpha ( 1 + \tan^2 \beta)

= ( 1 + \tan^2 \beta) ( 1 + \tan^2 \alpha)

= \sec^2 \alpha \sec^2 \beta =  RHS. Hence Proved.

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Question 14: \frac{(1+\cot x + \tan x)(\sin x - \cos x)}{\sec^3 x - \mathrm{cosec}^3 x} = \sin^2 x \cos^2 x

Answer:

LHS = \frac{(1+\cot x + \tan x)(\sin x - \cos x)}{\sec^3 x - \mathrm{cosec}^3 x}

= \frac{(1+\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x})(\sin x - \cos x)}{\frac{1}{\cos^3 x} - \frac{1}{\sin^3 x}}

= \frac{(\sin x \cos x + \cos^2 x +\sin^2 x)(\sin x - \cos x)}{\sin x \cos x ( \frac{\sin^3 x - \cos^3 x}{\sin^3 x \cos^3 x} )}

= \frac{(\sin x \cos x + 1)(\sin x - \cos x)(\sin^2 x \cos^2 x)}{\sin^3 x - \cos^3 x}

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

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

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

= \sin^2 x \cos^2 x

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Question 15: \frac{2 \sin x \cos x - \cos x}{1- \sin x + \sin^2 x-\cos^2 x} = \cot x

Answer:

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

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

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

= \cot x = RHS. Hence proved.

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Question 16: \cos x( \tan x + 2)(2 \tan x + 1) = 2 \sec x + 5 \sin x

Answer:

LHS = \cos x( \tan x + 2)(2 \tan x + 1)

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

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

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

= \frac{2  + 5 \sin x \cos x }{\cos x}

= 2 \sec x + 5 \sin x = RHS

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Question 17: If a = \frac{2 \sin x}{1 + \cos x + \sin x} , then prove that \frac{1 - \cos x + \sin x}{1+ \sin x} is also equal to a

Answer:

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

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

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

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

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

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

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

= \frac{ 1 - \cos x + \sin x}{\sin x +1 } = RHS. Hence proved.

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Question 18: If  \sin x = \frac{a^2 - b^2}{a^2 + b^2} , find the values of \tan x , \sec x and \mathrm{cosec} x .

Answer:

Given: \sin x = \frac{a^2 - b^2}{a^2 + b^2}

\cos x = \sqrt{1 - \sin^2 x}

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

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

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

= \frac{2ab}{a^2 + b^2}

\therefore \tan x = \frac{\sin x}{\cos x} = \frac{(a^2-b^2)(a^2+b^2)}{(a^2+b^2) \times 2ab} = \frac{a^2 - b^2}{2ab}

\sec x = \frac{1}{\cos x} = \frac{a^2+b^2}{2ab}

\mathrm{cosec} = \frac{1}{\sin x} = \frac{a^2 + b^2}{a^2 - b^2}

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Question 19: If \tan x = \frac{b}{a} , then find the value of \sqrt{\frac{a+b}{a-b}} + \sqrt{\frac{a-b}{a+b}}

Answer:

Given \tan x = \frac{b}{a}

\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 - (\frac{b}{a})}}    = \frac{2}{\sqrt{1 - \tan^2 x}}    = \frac{2 \cos x}{\sqrt{\cos^2 x - \sin^2 x}}

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Question 20: If \tan x = \frac{a}{b} , show that \frac{a \sin x - b \cos x}{a \sin x + b \cos x} = \frac{a^2 - b^2}{a^2+ b^2}

Answer:

LHS = \frac{a \sin x - b \cos x}{a \sin x + b \cos x}

= \frac{a \tan x - b}{a \tan x + b}    = \frac{a (\frac{a}{b}) - b}{a (\frac{a}{b}) + b}    = \frac{a^2 - b^2}{a^2 + b^2} = RHS. Hence proved.

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Question 21: If \mathrm{cosec} x - \sin x = a^3, \sec x - \cos x = b^3 , then prove that a^2b^2(a^2+b^2) = 1 .

Answer:

Given \mathrm{cosec} x - \sin x = a^3

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

Similarly given \sec x - \cos x = b^3

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

Therefore a^2 b^2 (a^2 + b^2) =a^4b^2 + a^2b^4

= \Big(  \frac{\cos^2 x}{\sin x} \Big)^{\frac{4}{3}} \Big(  \frac{\sin^2 x}{\cos x} \Big)^{\frac{2}{3}} + \Big(  \frac{\cos^2 x}{\sin x} \Big)^{\frac{2}{3}} \Big(  \frac{\sin^2 x}{\cos x} \Big)^{\frac{4}{3}}

= \frac{\cos^\frac{8}{3} x \sin^\frac{4}{3} x}{\sin^\frac{4}{3} x \cos^\frac{2}{3} x} + \frac{\cos^\frac{4}{3} x \sin^\frac{8}{3} x}{\sin^\frac{2}{3} x \cos^\frac{4}{3} x}

= \cos^\frac{4}{3} x \sin^\frac{4}{3} x \Bigg[  \frac{\cos^\frac{4}{3} x }{\sin^\frac{4}{3} x \cos^\frac{2}{3} x} +  \frac{\sin^\frac{4}{3} x }{\sin^\frac{2}{3} x \cos^\frac{4}{3} x}  \Bigg]

= \frac{\cos^\frac{4}{3} x \sin^\frac{4}{3} x}{\sin^\frac{2}{3} x \cos^\frac{2}{3} x}  \Bigg[  \frac{\cos^\frac{4}{3} x }{\sin^\frac{2}{3} x} +  \frac{\sin^\frac{4}{3} x }{ \cos^\frac{2}{3} x}  \Bigg]

= \frac{\cos^\frac{4}{3} x \sin^\frac{4}{3} x}{\sin^\frac{2}{3} x \cos^\frac{2}{3} x}  \Bigg[  \frac{ \cos^\frac{6}{3} x + \sin^\frac{6}{3} x }{ \sin^\frac{2}{3} x \cos^\frac{2}{3} x}  \Bigg]

= \frac{\cos^\frac{4}{3} x \sin^\frac{4}{3} x}{\sin^\frac{2}{3} x \cos^\frac{2}{3} x} \Bigg[  \frac{ \cos^2 x + \sin^2 x }{ \sin^\frac{2}{3} x \cos^\frac{2}{3} x} \Bigg]

= \frac{\cos^\frac{4}{3} x \sin^\frac{4}{3} x}{\sin^\frac{2}{3} x \cos^\frac{2}{3} x}  \Bigg[  \frac{ 1 }{ \sin^\frac{2}{3} x \cos^\frac{2}{3} x}  \Bigg]

= \frac{\cos^\frac{4}{3} x \sin^\frac{4}{3} x}{\sin^\frac{4}{3} x \cos^\frac{4}{3} x} = 1 = RHS.

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Question 22: If \cot x (1+ \sin x)= 4m and \cot x(1-\sin x) = 4n , then prove that (m^2 - n^2)^2 = mn .

Answer:

Given \cot x (1+ \sin x)= 4m    \Rightarrow m = \frac{\cot x(1 + \sin x)}{4}

Also \cot x(1-\sin x) = 4n    \Rightarrow n = \frac{\cot x(1 - \sin x)}{4}

LHS = (m^2 - n^2)^2

= \Big[ \frac{\cot^2 x(1 + \sin x)^2}{16} - \frac{\cot^2 x(1 - \sin x)^2}{16} \Big]^2

= \frac{\cot^4 x}{256} \Big[ (1 + \sin x)^2 - (1 - \sin x)^2 \Big]^2

= \frac{\cot^4 x}{256} \Big[ 1 + \sin^2 x + 2 \sin x - 1 - \sin^2 x + 2 \sin x \Big]^2

= \frac{\cot^4 x}{256} \Big[  4\sin x  \Big]^2

= \frac{\cot^4 x}{256} \times 16 \sin^2 x

= \frac{\cot^4 x}{16} \sin^2 x

= \frac{\cot^2 x \cos^2 x}{16}

RHS = mn = \frac{\cot x(1 + \sin x)}{4} . \frac{\cot x(1 - \sin x)}{4} = \frac{\cot^2 x \cos^2 x}{16}

Therefore LHS = RHS. Hence proved.

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Question 23: If \sin x + \cos x = m , then prove that                                                 \sin^6 x + \cos^6 x = \frac{4-3(m^2 - 1)^2}{4} , where m^2 \leq 2 .

Answer:

Given \sin x + \cos x = m

RHS = \frac{4-3(m^2 - 1)^2}{4}

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

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

= \frac{4-3(2 \sin x \cos x)^2}{4}

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

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

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

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

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

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

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

Therefore LHS = RHS. Hence proved.

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Question 24: If a = \sec x - \tan x and b = \mathrm{cosec} x + \cot x , then show that ab + a - b +  1 = 0  

Answer:

Given: a = \sec x - \tan x = \frac{1-\sin x}{\cos x}

b = \mathrm{cosec} x + \cot x = \frac{1+ \cos x}{\sin x}

LHS = ab + a - b +  1

= \frac{1-\sin x}{\cos x} . \frac{1+ \cos x}{\sin x} + \frac{1-\sin x}{\cos x} - \frac{1+ \cos x}{\sin x} + 1

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

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

= \frac{1 - 1}{\sin x \cos x} = 0= RHS. Hence proved.

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Question 25: Prove that: \Big|\sqrt{\frac{1-\sin x}{1+ \sin x}} + \sqrt{\frac{1+\sin x}{1- \sin x}} \Big| = - \frac{2}{\cos x} , where \frac{\pi}{2} < x < \pi .

Answer:

LHS = \Big|\sqrt{\frac{1-\sin x}{1+ \sin x}} + \sqrt{\frac{1+\sin x}{1- \sin x}} \Big|

= \Bigg| \frac{(1-\sin x) + (1 + \sin x)}{\sqrt{1 - \sin^2 x}} \Bigg| = \Bigg| \frac{2}{\cos x} \Bigg | = \frac{-2}{\cos x} since \frac{\pi}{2} < x < \pi .

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Question 26: If T_n = \sin^n x + \cos^n x , prove that                                                               i) \frac{T_3-T_5}{T_1} = \frac{T_5-T_7}{T_3}     ii) 2T_6-3T_4+1 = 0

Answer:

Given T_n = \sin^n x + \cos^n x

i) LHS = \frac{T_3-T_5}{T_1}

= \frac{ ( \sin^3 x + \cos^3 x) - ( \sin^5 x + \cos^5 x)}{ \sin x + \cos x}

= \frac{\sin^3 x ( 1 - \sin^2 x ) + \cos^3 x ( 1 - \cos^2 x)}{ \sin x + \cos x}

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

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

= \sin^2 x \cos^2 x

RHS = \frac{T_5-T_7}{T_3}

= \frac{ ( \sin^5 x + \cos^5 x) - ( \sin^7 x + \cos^7 x)}{ \sin^3 x + \cos^3 x}

= \frac{\sin^5 x ( 1 - \sin^2 x ) + \cos^5 x ( 1 - \cos^2 x)}{ \sin^3 x + \cos^3 x}

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

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

= \sin^2 x \cos^2 x

Therefore LHS = RHS. Hence proved.

ii) LHS = 2T_6-3T_4+1

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

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

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

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

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

= 2 [ 1 - 3 \sin^2 x . \cos^2 x  ] - 3[ 1 - 2 \sin^2 x . \cos^2 x ] + 1

=2 - 6 \sin^2 x . \cos^2 x- 3 + 6 \sin^2 x . \cos^2 x + 1

= 0 .  Hence proved.