Prove the following identities (1-16) :

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$= \sin^2 x \cos^2 x$

$\\$

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

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

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 $a$

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 $\mathrm{cosec} x$.

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

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

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

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

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

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

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

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

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

$\\$

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

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

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 $a^2b^2(a^2+b^2) = 1$.

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 $(m^2 - n^2)^2 = mn$.

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 $m^2 \leq 2$.

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 $ab + a - b + 1 = 0$

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 $\frac{\pi}{2}$ $< x < \pi$.

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) $2T_6-3T_4+1 = 0$

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.