Inverse Trigonometric Functions

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Please consider the following

$ta{n}^{-1}\frac{x}{y}-ta{n}^{-1}\frac{x-y}{x+y}\left\{-ta{n}^{-1}x-ta{n}^{-1}y=\frac{x-y}{1+xy}\right\}$

$=ta{n}^{-1}\left\{\frac{\left(\frac{x}{y}\right)-\left(\frac{x-y}{x+y}\right)}{1+\frac{x}{y}·\left(\frac{x-y}{x+y}\right)}\right\}$

$=ta{n}^{-1}\left\{\frac{\frac{x(x+y)-(x-y)·y}{y(x+y)}}{\frac{y(x+y)+x(x-y)}{y(x+y)}}\right\}$

$=ta{n}^{-1}\left(\frac{x^2+xy-xy+y^2}{xy+y^2-x^2-xy}\right)$

$=ta{n}^{-1}\frac{x^2+y^2}{x^2+y^2}$

$=ta{n}^{-1}1$

$=ta{n}^{-1}\left(tan\frac{\pi }{4}\right)$

$=\frac{\pi }{4}.$

Option C is correct.

$si{n}^{-1}(1-x)-2si{n}^{-1}x=\frac{\pi }{2}\to (1)$

(M) Let $x=sin\theta .Then,\theta =si{n}^{-1}x.$

Putting this in $q^n(1)$ we get

$si{n}^{-1}(1-x)-2·\theta =\frac{\pi }{2}$

$\Rightarrow si{n}^{-1}(1-x)=\frac{\pi }{2}+2\theta$

$\Rightarrow 1-x=sin\left(\frac{\pi }{2}+2\theta \right)\left\{sin\left(\frac{\pi }{2}+x\right)=cosx\right\}$

$\Rightarrow 1-x=cos2\theta$

$\Rightarrow 1-x=1-2si{n}^2\theta ·\left\{cos2x=1-2si{n}^{2}x\right\}$

$\Rightarrow 1-x=1-2x^2·\{sin\theta =x\}$

$\begin{array}{l}\Rightarrow 2x^2-x=0\\ \Rightarrow x(2x-1)=0\end{array}$

$\Rightarrow so,x=0x2x-1=0x2x=1\to x=\frac{1}{2}.$

Putting $x=0$ in qⁿ (1) .

L.H.S $=si{n}^{-1}(1-0)-2si{n}^{-1}0=si{n}^{-1}sin\frac{x}{2}-0=\frac{\pi }{2}=R.H.S.$

$x=\frac{1}{2}\mathrm{q(1)}$

$L.H.S=si{n}^{-1}\left(1-\frac{1}{2}\right)-2si{n}^{-1}\frac{1}{2}=si{n}^{-1}\left(\frac{1}{2}\right)-2si{n}^{-1}\frac{1}{2}$

$=si{n}^{-1}\frac{1}{2}-2si{n}^{-1}\frac{1}{2}$

$=-si{n}^{-1}\frac{1}{2}=-si{n}^{-1}\left(sin\frac{\pi }{6}\right)=-\frac{\pi }{6}$ $\ne$

So, $=0.$

Option (c) is correct.