Prove the following identities (1-16) :

$\displaystyle \text{Question 1: } \sec^4 x - \sec^2 x = \tan^4 x + \tan^2 x$

$\displaystyle \text{LHS } = \sec^4 x - \sec^2 x$

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

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

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

$\displaystyle \\$

$\displaystyle \text{Question 2: } \sin^6 x + \cos^6 x = 1 - 3 \sin^2 x \cos^2 x$

$\displaystyle \text{LHS } = \sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3$

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

$\displaystyle = 1 =$ RHS. Hence proved.

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

$\displaystyle = \sin x = \text{RHS. Hence proved.}$

$\displaystyle \\$

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

$\displaystyle = 2 = \text{RHS. Hence proved.}$

$\displaystyle \\$

$\displaystyle \text{Question 8: } (\sec x \sec y + \tan x \tan y )^2 - (\sec x \tan y + \tan x \sec y)^2 = 1$

$\displaystyle \text{LHS } = (\sec x \sec y + \tan x \tan y )^2 - (\sec x \tan y + \tan x \sec y)^2$

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

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

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

$\displaystyle = sec^2 x + \tan^2 x = 1 = \text{RHS. Hence proved.}$

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

$\displaystyle = \sin x \cos x = \text{ RHS. Hence Proved. }$

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

$\displaystyle \text{Question 13: } (1 + \tan \alpha \tan \beta)^2 + (\tan \alpha - \tan \beta)^2 = \sec^2 \alpha \sec^2 \beta$

$\displaystyle \text{LHS } = (1 + \tan \alpha \tan \beta)^2 + (\tan \alpha - \tan \beta)^2$

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

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

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

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

$\displaystyle = \sec^2 \alpha \sec^2 \beta = \text{ RHS. Hence Proved. }$

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

$\displaystyle = \cot x = \text{ RHS. Hence Proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 16: } \cos x( \tan x + 2)(2 \tan x + 1) = 2 \sec x + 5 \sin x$

$\displaystyle \text{LHS } = \cos x( \tan x + 2)(2 \tan x + 1)$

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

$\displaystyle = \frac{ 1 - \cos x + \sin x}{\sin x +1 } = \text{ RHS. Hence Proved. }$

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \\$

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

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

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

$\displaystyle \\$

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

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

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

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

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

Therefore $\displaystyle a^2 b^2 (a^2 + b^2) =a^4b^2 + a^2b^4$

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

$\displaystyle \text{LHS } = (m^2 - n^2)^2$

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

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

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

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

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

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

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

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

Therefore $\displaystyle \text{LHS } = \text{ RHS. Hence Proved. }$

$\displaystyle \\$

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

Given $\displaystyle \sin x + \cos x = m$

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

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

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

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

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

$\displaystyle \text{LHS } = \sin^6 x + \cos^6 x$

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

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

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

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

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

Therefore $\displaystyle \text{LHS } = \text{ RHS. Hence Proved. }$

$\displaystyle \\$

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

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

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

$\displaystyle \text{LHS } = ab + a - b + 1$

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

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

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

$\displaystyle = \frac{1 - 1}{\sin x \cos x} = 0= \text{ RHS. Hence Proved. }$

$\displaystyle \\$

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

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

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

$\displaystyle \\$

$\displaystyle \text{Question 26: If } T_n = \sin^n x + \cos^n x \text{, prove that }$

$\displaystyle \text{i) } \frac{T_3-T_5}{T_1} = \frac{T_5-T_7}{T_3} \hspace{1.0cm} \text{ii) } 2T_6-3T_4+1 = 0$

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

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

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

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

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

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

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

$\displaystyle \text{RHS } = \frac{T_5-T_7}{T_3}$

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

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

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

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

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

Therefore $\displaystyle \text{LHS } = \text{ RHS. Hence Proved. }$

$\displaystyle \text{ii) LHS } = 2T_6-3T_4+1$

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

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

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

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

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

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

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

$\displaystyle = 0 \text{ RHS. Hence Proved. }$