Solveis equal to

(A) (B) (C) (D)

Q. Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.
${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$
$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.
If $0<x<1$, then
$\sqrt{1+x^2}{[{\left\{xcos\right({cot}^{-1}x)+sin({cot}^{-1}x\left)\right\}}^2-1]}^{1/2}$ is equal to
(a) $\frac{x}{\sqrt{1+x^2}}$
(b) $x$
(c) $x\sqrt{1+x^2}$
(d) $\sqrt{1+x^2}$

Q. Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.
${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$
$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.
(a) $3{sin}^{-1}\frac{2x}{1+x^2}-4{cos}^{-1}\frac{1-x^2}{1+x^2}+2{tan}^{-1}\frac{2x}{1-x^2}=\frac{\pi }{3}$
(b) $sin\left[\right(1/5\left){cos}^{-1}x\right]=1$
(c) ${tan}^{-1}\sqrt{x^2+x}+{sin}^{-1}\sqrt{x^2+x+1}=\frac{\pi }{2}$

Q. Solve for $x$:${\sin }^{-1}\frac{x}{\sqrt{1+x^2}}+{\sin }^{-1}\frac{1}{\sqrt{1+x^2}}={\sin }^{-1}\left(\frac{1+x}{\sqrt{1+x^2}}\right)$

Q. The value of $sin\left(co{t}^{-1}\right(cos\left(ta{n}^{-1}x\right)\left)\right)$ is equal to

Q. Prove the following
$\left(1\right){\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=2{\tan }^{-1}x,|x|\le 1$
$\left(2\right){\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2{\tan }^{-1}x,x\ge 0$
$\left(3\right){\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=2{\tan }^{-1}x,-1<x<1$