Class 11: Trigonometric Functions

Prove the following identities (1-16) :

$\displaystyle \textbf{Question 1: }\sec^4x-\sec^2x=\tan^4x+\tan^2x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\sec^4x-\sec^2x$
$\displaystyle =\sec^2x(\sec^2x-1)$
$\displaystyle =(1+\tan^2x)\tan^2x$
$\displaystyle =\tan^2x+\tan^4x$
$\displaystyle =\tan^4x+\tan^2x=\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\sin^6x+\cos^6x=1-3\sin^2x\cos^2x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\sin^6x+\cos^6x$
$\displaystyle =(\sin^2x)^3+(\cos^2x)^3$
$\displaystyle =(\sin^2x+\cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x)$
$\displaystyle =\sin^4x+\cos^4x-\sin^2x\cos^2x$
$\displaystyle =(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x-\sin^2x\cos^2x$
$\displaystyle =1-3\sin^2x\cos^2x=\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }(\mathrm{cosec}x-\sin x)(\sec x-\cos x)(\tan x+\cot x)=1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=(\mathrm{cosec}x-\sin x)(\sec x-\cos x)(\tan x+\cot x)$
$\displaystyle =\left(\frac{1}{\sin x}-\sin x\right)\left(\frac{1}{\cos x}-\cos x\right)$
$\displaystyle \qquad \times\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)$
$\displaystyle =\frac{1-\sin^2x}{\sin x}\cdot\frac{1-\cos^2x}{\cos x}\cdot$
$\displaystyle \qquad\frac{\sin^2x+\cos^2x}{\sin x\cos x}$
$\displaystyle =\frac{\cos^2x}{\sin x}\cdot\frac{\sin^2x}{\cos x}\cdot\frac{1}{\sin x\cos x}$
$\displaystyle =1=\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\mathrm{cosec}x(\sec x-1)-\cot x(1-\cos x)=\tan x-\sin x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\mathrm{cosec}x(\sec x-1)-\cot x(1-\cos x)$
$\displaystyle =\frac{1}{\sin x}\left(\frac{1}{\cos x}-1\right)-\frac{\cos x}{\sin x}(1-\cos x)$
$\displaystyle =\frac{1-\cos x}{\sin x\cos x}-\frac{(1-\cos x)\cos x}{\sin x}$
$\displaystyle =\frac{1-\cos x}{\sin x\cos x}-\frac{(1-\cos x)\cos^2x}{\sin x\cos x}$
$\displaystyle =\frac{(1-\cos x)(1-\cos^2x)}{\sin x\cos x}$
$\displaystyle =\frac{(1-\cos x)\sin^2x}{\sin x\cos x}$
$\displaystyle =\frac{\sin x}{\cos x}(1-\cos x)$
$\displaystyle =\tan x-\sin x=\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\frac{1-\sin x\cos x}{\cos x(\sec x-\mathrm{cosec}x)}\cdot$
$\displaystyle \frac{\sin^2x-\cos^2x}{\sin^3x+\cos^3x}=\sin x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{1-\sin x\cos x}{\cos x(\sec x-\mathrm{cosec}x)}\cdot$
$\displaystyle \frac{\sin^2x-\cos^2x}{\sin^3x+\cos^3x}$
$\displaystyle =\frac{1-\sin x\cos x}{\cos x\left(\frac{1}{\cos x}-\frac{1}{\sin x}\right)}\cdot$
$\displaystyle \frac{\sin^2x-\cos^2x}{\sin^3x+\cos^3x}$
$\displaystyle =\frac{1-\sin x\cos x}{\cos x\left(\frac{\sin x-\cos x}{\sin x\cos x}\right)}\cdot$
$\displaystyle \frac{(\sin x-\cos x)(\sin x+\cos x)}{(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)}$
$\displaystyle =\frac{\sin x(1-\sin x\cos x)}{\sin x-\cos x}\cdot$
$\displaystyle \frac{\sin x-\cos x}{1-\sin x\cos x}$
$\displaystyle =\sin x=\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x}$
$\displaystyle =\sec x\,\mathrm{cosec}x+1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x}$
$\displaystyle =\frac{\sin x}{\cos x\left(1-\frac{\cos x}{\sin x}\right)}+$
$\displaystyle \frac{\cos x}{\sin x\left(1-\frac{\sin x}{\cos x}\right)}$
$\displaystyle =\frac{\sin^2x}{\cos x(\sin x-\cos x)}+$
$\displaystyle \frac{\cos^2x}{\sin x(\cos x-\sin x)}$
$\displaystyle =\frac{\sin^3x-\cos^3x}{\sin x\cos x(\sin x-\cos x)}$
$\displaystyle =\frac{(\sin x-\cos x)(\sin^2x+\sin x\cos x+\cos^2x)}{\sin x\cos x(\sin x-\cos x)}$
$\displaystyle =\frac{1+\sin x\cos x}{\sin x\cos x}$
$\displaystyle =\frac{1}{\sin x\cos x}+1$
$\displaystyle =\sec x\,\mathrm{cosec}x+1=\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\frac{\sin^3x+\cos^3x}{\sin x+\cos x}+\frac{\sin^3x-\cos^3x}{\sin x-\cos x}=2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{\sin^3x+\cos^3x}{\sin x+\cos x}+\frac{\sin^3x-\cos^3x}{\sin x-\cos x}$
$\displaystyle =\frac{(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)}{\sin x+\cos x}$
$\displaystyle \qquad+\frac{(\sin x-\cos x)(\sin^2x+\sin x\cos x+\cos^2x)}{\sin x-\cos x}$
$\displaystyle =1-\sin x\cos x+1+\sin x\cos x$
$\displaystyle =2=\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }(\sec x\sec y+\tan x\tan y)^2$
$\displaystyle -(\sec x\tan y+\tan x\sec y)^2=1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=(\sec x\sec y+\tan x\tan y)^2$
$\displaystyle \qquad-(\sec x\tan y+\tan x\sec y)^2$
$\displaystyle =\sec^2x\sec^2y+\tan^2x\tan^2y+2\sec x\sec y\tan x\tan y$
$\displaystyle \qquad-\sec^2x\tan^2y-\tan^2x\sec^2y-2\sec x\tan y\tan x\sec y$
$\displaystyle =\sec^2x\sec^2y+\tan^2x\tan^2y-\sec^2x\tan^2y-\tan^2x\sec^2y$
$\displaystyle =\sec^2x(\sec^2y-\tan^2y)+\tan^2x(\tan^2y-\sec^2y)$
$\displaystyle =\sec^2x-\tan^2x$
$\displaystyle =1=\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\frac{\cos x}{1-\sin x}=\frac{1+\cos x+\sin x}{1+\cos x-\sin x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{RHS}=\frac{1+\cos x+\sin x}{1+\cos x-\sin x}$
$\displaystyle =\frac{(1+\cos x)+\sin x}{(1+\cos x)-\sin x}\cdot$
$\displaystyle \frac{(1+\cos x)+\sin x}{(1+\cos x)+\sin x}$
$\displaystyle =\frac{(1+\cos x+\sin x)^2}{(1+\cos x)^2-\sin^2x}$
$\displaystyle =\frac{1+\cos^2x+\sin^2x+2\cos x+2\sin x+2\sin x\cos x}{1+\cos^2x+2\cos x-\sin^2x}$
$\displaystyle \qquad{1+\cos^2x+2\cos x-\sin^2x}$
$\displaystyle =\frac{2+2\cos x+2\sin x+2\sin x\cos x}{2\cos^2x+2\cos x}$
$\displaystyle =\frac{2(1+\cos x)(1+\sin x)}{2\cos x(1+\cos x)}$
$\displaystyle =\frac{1+\sin x}{\cos x}$
$\displaystyle =\frac{1+\sin x}{\cos x}\cdot\frac{1-\sin x}{1-\sin x}$
$\displaystyle =\frac{1-\sin^2x}{\cos x(1-\sin x)}$
$\displaystyle =\frac{\cos^2x}{\cos x(1-\sin x)}$
$\displaystyle =\frac{\cos x}{1-\sin x}=\text{LHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\frac{\tan^3x}{1+\tan^2x}+\frac{\cot^3x}{1+\cot^2x}$
$\displaystyle =\frac{1-2\sin^2x\cos^2x}{\sin x\cos x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{\tan^3x}{1+\tan^2x}+\frac{\cot^3x}{1+\cot^2x}$
$\displaystyle =\frac{\tan^3x}{\sec^2x}+\frac{\cot^3x}{\mathrm{cosec}^2x}$
$\displaystyle =\frac{\sin^3x}{\cos^3x}\cdot\cos^2x+$
$\displaystyle \frac{\cos^3x}{\sin^3x}\cdot\sin^2x$
$\displaystyle =\frac{\sin^3x}{\cos x}+\frac{\cos^3x}{\sin x}$
$\displaystyle =\frac{\sin^4x+\cos^4x}{\sin x\cos x}$
$\displaystyle =\frac{(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x}{\sin x\cos x}$
$\displaystyle =\frac{1-2\sin^2x\cos^2x}{\sin x\cos x}=\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }1-\frac{\sin^2x}{1+\cot x}-\frac{\cos^2x}{1+\tan x}$
$\displaystyle =\sin x\cos x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=1-\frac{\sin^2x}{1+\cot x}-\frac{\cos^2x}{1+\tan x}$
$\displaystyle =1-\frac{\sin^2x}{1+\frac{\cos x}{\sin x}}-\frac{\cos^2x}{1+\frac{\sin x}{\cos x}}$
$\displaystyle =1-\frac{\sin^3x}{\sin x+\cos x}-\frac{\cos^3x}{\sin x+\cos x}$
$\displaystyle =\frac{\sin x+\cos x-\sin^3x-\cos^3x}{\sin x+\cos x}$
$\displaystyle =\frac{\sin x(1-\sin^2x)+\cos x(1-\cos^2x)}{\sin x+\cos x}$
$\displaystyle =\frac{\sin x\cos^2x+\cos x\sin^2x}{\sin x+\cos x}$
$\displaystyle =\frac{\sin x\cos x(\sin x+\cos x)}{\sin x+\cos x}$
$\displaystyle =\sin x\cos x=\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\left(\frac{1}{\sec^2x-\cos^2x}+\frac{1}{\mathrm{cosec}^2x-\sin^2x}\right)$
$\displaystyle \sin^2x\cos^2x=\frac{1-\sin^2x\cos^2x}{2+\sin^2x\cos^2x}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\left(\frac{1}{\sec^2x-\cos^2x}+\frac{1}{\mathrm{cosec}^2x-\sin^2x}\right)\sin^2x\cos^2x$
$\displaystyle =\left(\frac{1}{\frac{1}{\cos^2x}-\cos^2x}+\frac{1}{\frac{1}{\sin^2x}-\sin^2x}\right)\sin^2x\cos^2x$
$\displaystyle =\left(\frac{\cos^2x}{1-\cos^4x}+\frac{\sin^2x}{1-\sin^4x}\right)\sin^2x\cos^2x$
$\displaystyle =\left(\frac{\cos^2x}{(1-\cos^2x)(1+\cos^2x)}+\frac{\sin^2x}{(1-\sin^2x)(1+\sin^2x)}\right)$
$\displaystyle \qquad\times\sin^2x\cos^2x$
$\displaystyle =\left(\frac{\cos^2x}{\sin^2x(1+\cos^2x)}+\frac{\sin^2x}{\cos^2x(1+\sin^2x)}\right)$
$\displaystyle \qquad\times\sin^2x\cos^2x$
$\displaystyle =\frac{\cos^4x(1+\sin^2x)+\sin^4x(1+\cos^2x)}{(1+\cos^2x)(1+\sin^2x)}$
$\displaystyle =\frac{\sin^4x+\cos^4x+\sin^2x\cos^2x(\sin^2x+\cos^2x)}{(1+\cos^2x)(1+\sin^2x)}$
$\displaystyle =\frac{(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x+\sin^2x\cos^2x}{1+\sin^2x+\cos^2x+\sin^2x\cos^2x}$
$\displaystyle =\frac{1-\sin^2x\cos^2x}{2+\sin^2x\cos^2x}$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }(1+\tan\alpha\tan\beta)^2+(\tan\alpha-\tan\beta)^2$
$\displaystyle =\sec^2\alpha\sec^2\beta$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=(1+\tan\alpha\tan\beta)^2+(\tan\alpha-\tan\beta)^2$
$\displaystyle =1+\tan^2\alpha\tan^2\beta+2\tan\alpha\tan\beta+\tan^2\alpha+\tan^2\beta$
$\displaystyle \qquad-2\tan\alpha\tan\beta$
$\displaystyle =1+\tan^2\alpha\tan^2\beta+\tan^2\alpha+\tan^2\beta$
$\displaystyle =1+\tan^2\beta+\tan^2\alpha(1+\tan^2\beta)$
$\displaystyle =(1+\tan^2\alpha)(1+\tan^2\beta)$
$\displaystyle =\sec^2\alpha\sec^2\beta=\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\frac{(1+\cot x+\tan x)(\sin x-\cos x)}{\sec^3x-\mathrm{cosec}^3x}$
$\displaystyle =\sin^2x\cos^2x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{(1+\cot x+\tan x)(\sin x-\cos x)}{\sec^3x-\mathrm{cosec}^3x}$
$\displaystyle =\frac{\left(1+\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}\right)(\sin x-\cos x)}{\frac{1}{\cos^3x}-\frac{1}{\sin^3x}}$
$\displaystyle =\frac{(\sin x\cos x+\cos^2x+\sin^2x)(\sin x-\cos x)}{\sin x\cos x\left(\frac{\sin^3x-\cos^3x}{\sin^3x\cos^3x}\right)}$
$\displaystyle =\frac{(1+\sin x\cos x)(\sin x-\cos x)\sin^2x\cos^2x}{\sin^3x-\cos^3x}$
$\displaystyle =\frac{(1+\sin x\cos x)(\sin x-\cos x)\sin^2x\cos^2x}{(\sin x-\cos x)(\sin^2x+\sin x\cos x+\cos^2x)}$
$\displaystyle =\frac{(1+\sin x\cos x)\sin^2x\cos^2x}{1+\sin x\cos x}$
$\displaystyle =\sin^2x\cos^2x=\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\frac{2\sin x\cos x-\cos x}{1-\sin x+\sin^2x-\cos^2x}=\cot x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{2\sin x\cos x-\cos x}{1-\sin x+\sin^2x-\cos^2x}$
$\displaystyle =\frac{\cos x(2\sin x-1)}{\sin^2x-\sin x+\sin^2x}$
$\displaystyle =\frac{\cos x(2\sin x-1)}{\sin x(2\sin x-1)}$
$\displaystyle =\frac{\cos x}{\sin x}$
$\displaystyle =\cot x=\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\cos x(\tan x+2)(2\tan x+1)=2\sec x+5\sin x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\cos x(\tan x+2)(2\tan x+1)$
$\displaystyle =\cos x\left(\frac{\sin x}{\cos x}+2\right)\left(\frac{2\sin x}{\cos x}+1\right)$
$\displaystyle =\frac{(\sin x+2\cos x)(2\sin x+\cos x)}{\cos x}$
$\displaystyle =\frac{2\sin^2x+5\sin x\cos x+2\cos^2x}{\cos x}$
$\displaystyle =\frac{2(\sin^2x+\cos^2x)+5\sin x\cos x}{\cos x}$
$\displaystyle =\frac{2+5\sin x\cos x}{\cos x}$
$\displaystyle =2\sec x+5\sin x=\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{If }a=\frac{2\sin x}{1+\cos x+\sin x},\text{ prove that}$
$\displaystyle \frac{1-\cos x+\sin x}{1+\sin x}\text{ is also equal to }a.$
$\displaystyle \text{Answer:}$
$\displaystyle a=\frac{2\sin x}{1+\cos x+\sin x}$
$\displaystyle =\frac{2\sin x}{1+\cos x+\sin x}\cdot$
$\displaystyle \frac{1-\cos x+\sin x}{1-\cos x+\sin x}$
$\displaystyle =\frac{2\sin x(1-\cos x+\sin x)}{(1+\sin x)^2-\cos^2x}$
$\displaystyle =\frac{2\sin x(1-\cos x+\sin x)}{1+\sin^2x+2\sin x-\cos^2x}$
$\displaystyle =\frac{2\sin x(1-\cos x+\sin x)}{2\sin^2x+2\sin x}$
$\displaystyle =\frac{2\sin x(1-\cos x+\sin x)}{2\sin x(1+\sin x)}$
$\displaystyle =\frac{1-\cos x+\sin x}{1+\sin x}$
$\displaystyle \therefore \frac{1-\cos x+\sin x}{1+\sin x}=a.$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{If }\sin x=\frac{a^2-b^2}{a^2+b^2},\text{ find the values of }\tan x,\sec x$
$\displaystyle \text{and }\mathrm{cosec}x,\text{ assuming that }x\text{ is acute and }a>b>0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\sin x=\frac{a^2-b^2}{a^2+b^2}$
$\displaystyle \cos x=\sqrt{1-\sin^2x}$
$\displaystyle =\sqrt{1-\left(\frac{a^2-b^2}{a^2+b^2}\right)^2}$
$\displaystyle =\sqrt{\frac{(a^2+b^2)^2-(a^2-b^2)^2}{(a^2+b^2)^2}}$
$\displaystyle =\sqrt{\frac{4a^2b^2}{(a^2+b^2)^2}}$
$\displaystyle =\frac{2ab}{a^2+b^2}$
$\displaystyle \therefore \tan x=\frac{\sin x}{\cos x}$
$\displaystyle =\frac{a^2-b^2}{a^2+b^2}\cdot\frac{a^2+b^2}{2ab}$
$\displaystyle =\frac{a^2-b^2}{2ab}$
$\displaystyle \sec x=\frac{1}{\cos x}=\frac{a^2+b^2}{2ab}$
$\displaystyle \mathrm{cosec}x=\frac{1}{\sin x}=\frac{a^2+b^2}{a^2-b^2}$
$\displaystyle \therefore \tan x=\frac{a^2-b^2}{2ab},\quad\sec x=\frac{a^2+b^2}{2ab}$
$\displaystyle \text{and }\mathrm{cosec}x=\frac{a^2+b^2}{a^2-b^2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{If }\tan x=\frac{b}{a},\text{ find the value of}$
$\displaystyle \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}},\text{ where }a>b>0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\tan x=\frac{b}{a}$
$\displaystyle \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$
$\displaystyle =\frac{a+b}{\sqrt{(a+b)(a-b)}}+\frac{a-b}{\sqrt{(a-b)(a+b)}}$
$\displaystyle =\frac{(a+b)+(a-b)}{\sqrt{a^2-b^2}}$
$\displaystyle =\frac{2a}{\sqrt{a^2-b^2}}$
$\displaystyle =\frac{2a}{a\sqrt{1-\left(\frac{b}{a}\right)^2}}$
$\displaystyle =\frac{2}{\sqrt{1-\tan^2x}}$
$\displaystyle =\frac{2\cos x}{\sqrt{\cos^2x-\sin^2x}}$
$\displaystyle \therefore \text{The required value is }\frac{2}{\sqrt{1-\tan^2x}}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{If }\tan x=\frac{a}{b},\text{ show that}$
$\displaystyle \frac{a\sin x-b\cos x}{a\sin x+b\cos x}=\frac{a^2-b^2}{a^2+b^2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{a\sin x-b\cos x}{a\sin x+b\cos x}$
$\displaystyle =\frac{a\tan x-b}{a\tan x+b}$
$\displaystyle =\frac{a\left(\frac{a}{b}\right)-b}{a\left(\frac{a}{b}\right)+b}$
$\displaystyle =\frac{\frac{a^2-b^2}{b}}{\frac{a^2+b^2}{b}}$
$\displaystyle =\frac{a^2-b^2}{a^2+b^2}=\text{RHS}$
$\displaystyle \therefore \text{The identity is proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{If }\mathrm{cosec}x-\sin x=a^3\text{ and }\sec x-\cos x=b^3,$
$\displaystyle \text{prove that }a^2b^2(a^2+b^2)=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\mathrm{cosec}x-\sin x=a^3$
$\displaystyle \frac{1}{\sin x}-\sin x=a^3$
$\displaystyle \frac{1-\sin^2x}{\sin x}=a^3$
$\displaystyle \therefore a^3=\frac{\cos^2x}{\sin x}$
$\displaystyle \text{Also, }\sec x-\cos x=b^3$
$\displaystyle \frac{1}{\cos x}-\cos x=b^3$
$\displaystyle \frac{1-\cos^2x}{\cos x}=b^3$
$\displaystyle \therefore b^3=\frac{\sin^2x}{\cos x}$
$\displaystyle (a^4b^2)^3=a^{12}b^6=(a^3)^4(b^3)^2$
$\displaystyle =\left(\frac{\cos^2x}{\sin x}\right)^4\left(\frac{\sin^2x}{\cos x}\right)^2$
$\displaystyle =\frac{\cos^8x}{\sin^4x}\cdot\frac{\sin^4x}{\cos^2x}$
$\displaystyle =\cos^6x$
$\displaystyle \therefore a^4b^2=\cos^2x$
$\displaystyle \text{Similarly,}$
$\displaystyle (a^2b^4)^3=a^6b^{12}=(a^3)^2(b^3)^4$
$\displaystyle =\left(\frac{\cos^2x}{\sin x}\right)^2\left(\frac{\sin^2x}{\cos x}\right)^4$
$\displaystyle =\frac{\cos^4x}{\sin^2x}\cdot\frac{\sin^8x}{\cos^4x}$
$\displaystyle =\sin^6x$
$\displaystyle \therefore a^2b^4=\sin^2x$
$\displaystyle a^2b^2(a^2+b^2)=a^4b^2+a^2b^4$
$\displaystyle =\cos^2x+\sin^2x$
$\displaystyle =1$
$\displaystyle \therefore a^2b^2(a^2+b^2)=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{If }\cot x(1+\sin x)=4m\text{ and }\cot x(1-\sin x)=4n,\text{ then}$
$\displaystyle \text{prove that }(m^2-n^2)^2=mn.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\cot x(1+\sin x)=4m$
$\displaystyle \therefore m=\frac{\cot x(1+\sin x)}{4}$
$\displaystyle \text{Also, }\cot x(1-\sin x)=4n$
$\displaystyle \therefore n=\frac{\cot x(1-\sin x)}{4}$
$\displaystyle \text{LHS}=(m^2-n^2)^2$
$\displaystyle =\left[\frac{\cot^2x(1+\sin x)^2}{16}-\frac{\cot^2x(1-\sin x)^2}{16}\right]^2$
$\displaystyle =\frac{\cot^4x}{256}\left[(1+\sin x)^2-(1-\sin x)^2\right]^2$
$\displaystyle =\frac{\cot^4x}{256}\left[4\sin x\right]^2$
$\displaystyle =\frac{\cot^4x}{16}\sin^2x$
$\displaystyle =\frac{\cot^2x}{16}\left(\cot^2x\sin^2x\right)$
$\displaystyle =\frac{\cot^2x\cos^2x}{16}$
$\displaystyle \text{RHS}=mn$
$\displaystyle =\frac{\cot x(1+\sin x)}{4}\times\frac{\cot x(1-\sin x)}{4}$
$\displaystyle =\frac{\cot^2x(1-\sin^2x)}{16}$
$\displaystyle =\frac{\cot^2x\cos^2x}{16}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{If }\sin x+\cos x=m,\text{ then prove that}$
$\displaystyle \sin^6x+\cos^6x=\frac{4-3(m^2-1)^2}{4},\text{ where }m^2\leq2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\sin x+\cos x=m$
$\displaystyle \text{RHS}=\frac{4-3(m^2-1)^2}{4}$
$\displaystyle =\frac{4-3\left[(\sin x+\cos x)^2-1\right]^2}{4}$
$\displaystyle =\frac{4-3\left(\sin^2x+\cos^2x+2\sin x\cos x-1\right)^2}{4}$
$\displaystyle =\frac{4-3(2\sin x\cos x)^2}{4}$
$\displaystyle =1-3\sin^2x\cos^2x$
$\displaystyle \text{LHS}=\sin^6x+\cos^6x$
$\displaystyle =(\sin^2x)^3+(\cos^2x)^3$
$\displaystyle =(\sin^2x+\cos^2x)(\sin^4x+\cos^4x-\sin^2x\cos^2x)$
$\displaystyle =\sin^4x+\cos^4x-\sin^2x\cos^2x$
$\displaystyle =(\sin^2x+\cos^2x)^2-3\sin^2x\cos^2x$
$\displaystyle =1-3\sin^2x\cos^2x$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{If }a=\sec x-\tan x\text{ and }b=\mathrm{cosec}\,x+\cot x,\text{ then show that}$
$\displaystyle ab+a-b+1=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }a=\sec x-\tan x$
$\displaystyle =\frac{1}{\cos x}-\frac{\sin x}{\cos x}=\frac{1-\sin x}{\cos x}$
$\displaystyle \text{Also, }b=\mathrm{cosec}\,x+\cot x$
$\displaystyle =\frac{1}{\sin x}+\frac{\cos x}{\sin x}=\frac{1+\cos x}{\sin x}$
$\displaystyle \text{LHS}=ab+a-b+1$
$\displaystyle =\frac{1-\sin x}{\cos x}\times\frac{1+\cos x}{\sin x}+\frac{1-\sin x}{\cos x}$
$\displaystyle \quad-\frac{1+\cos x}{\sin x}+1$
$\displaystyle =\frac{(1-\sin x)(1+\cos x)+\sin x(1-\sin x)}{\sin x\cos x}$
$\displaystyle \quad+\frac{-\cos x(1+\cos x)+\sin x\cos x}{\sin x\cos x}$
$\displaystyle =\frac{1-\sin x+\cos x-\sin x\cos x+\sin x-\sin^2x}{\sin x\cos x}$
$\displaystyle \quad+\frac{-\cos x-\cos^2x+\sin x\cos x}{\sin x\cos x}$
$\displaystyle =\frac{1-\sin^2x-\cos^2x}{\sin x\cos x}$
$\displaystyle =\frac{1-1}{\sin x\cos x}=0$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{Prove that}$
$\displaystyle \left|\sqrt{\frac{1-\sin x}{1+\sin x}}+\sqrt{\frac{1+\sin x}{1-\sin x}}\right|=-\frac{2}{\cos x},$
$\displaystyle \text{where }\frac{\pi}{2}<x<\pi.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\left|\sqrt{\frac{1-\sin x}{1+\sin x}}+\sqrt{\frac{1+\sin x}{1-\sin x}}\right|$
$\displaystyle =\left|\frac{1-\sin x}{\sqrt{1-\sin^2x}}+\frac{1+\sin x}{\sqrt{1-\sin^2x}}\right|$
$\displaystyle =\left|\frac{2}{\sqrt{1-\sin^2x}}\right|$
$\displaystyle =\left|\frac{2}{\sqrt{\cos^2x}}\right|$
$\displaystyle =\left|\frac{2}{|\cos x|}\right|$
$\displaystyle =\frac{2}{|\cos x|}$
$\displaystyle \text{Since }\frac{\pi}{2}<x<\pi,\text{ we have }\cos x<0.$
$\displaystyle \therefore |\cos x|=-\cos x$
$\displaystyle \therefore \text{LHS}=\frac{2}{-\cos x}=-\frac{2}{\cos x}=\text{RHS}$
$\displaystyle \therefore \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{If }T_n=\sin^n x+\cos^n x,\text{ prove that:}$
$\displaystyle \text{(i) }\frac{T_3-T_5}{T_1}=\frac{T_5-T_7}{T_3},\text{ provided }T_1T_3\neq0,$
$\displaystyle \text{(ii) }2T_6-3T_4+1=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }T_n=\sin^n x+\cos^n x.$
$\displaystyle \text{(i) LHS}=\frac{T_3-T_5}{T_1}$
$\displaystyle =\frac{\sin^3x+\cos^3x-\sin^5x-\cos^5x}{\sin x+\cos x}$
$\displaystyle =\frac{\sin^3x(1-\sin^2x)+\cos^3x(1-\cos^2x)}{\sin x+\cos x}$
$\displaystyle =\frac{\sin^3x\cos^2x+\cos^3x\sin^2x}{\sin x+\cos x}$
$\displaystyle =\frac{\sin^2x\cos^2x(\sin x+\cos x)}{\sin x+\cos x}$
$\displaystyle =\sin^2x\cos^2x.$
$\displaystyle \text{RHS}=\frac{T_5-T_7}{T_3}$
$\displaystyle =\frac{\sin^5x+\cos^5x-\sin^7x-\cos^7x}{\sin^3x+\cos^3x}$
$\displaystyle =\frac{\sin^5x(1-\sin^2x)+\cos^5x(1-\cos^2x)}{\sin^3x+\cos^3x}$
$\displaystyle =\frac{\sin^5x\cos^2x+\cos^5x\sin^2x}{\sin^3x+\cos^3x}$
$\displaystyle =\frac{\sin^2x\cos^2x(\sin^3x+\cos^3x)}{\sin^3x+\cos^3x}$
$\displaystyle =\sin^2x\cos^2x.$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \text{(ii) LHS}=2T_6-3T_4+1$
$\displaystyle =2(\sin^6x+\cos^6x)-3(\sin^4x+\cos^4x)+1.$
$\displaystyle \text{Now, }\sin^6x+\cos^6x$
$\displaystyle =(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)$
$\displaystyle =1-3\sin^2x\cos^2x.$
$\displaystyle \text{Also, }\sin^4x+\cos^4x$
$\displaystyle =(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$
$\displaystyle =1-2\sin^2x\cos^2x.$
$\displaystyle \therefore \text{LHS}=2(1-3\sin^2x\cos^2x)-3(1-2\sin^2x\cos^2x)+1$
$\displaystyle =2-6\sin^2x\cos^2x-3+6\sin^2x\cos^2x+1$
$\displaystyle =0=\text{RHS. Hence proved.}$
$\displaystyle \\$