Question

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) $\sqrt{1+x^2}$

Answer 1

(c) Let ${cot}^{-1}x=\theta$ $\therefore$ $cot\theta =x$
$\Rightarrow$ $cosec\theta =\sqrt{1+x^2}$ or $sin\theta =\frac{1}{\sqrt{1+x^2}}$
and $cos\theta =\frac{x}{\sqrt{1+x^2}}$
$\therefore$ $f\left(x\right)=\sqrt{1+x^2}{[{\{xcos\theta +sin\theta \}}^2-1]}^{1/2}$
$\sqrt{1+x^2}{[{\{x.\frac{x}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}\}}^2-1]}^{-1/2}$
$\sqrt{1+x^2}{[1+x^2-1]}^{1/2}=x\sqrt{1+x^2}\Rightarrow$ (c)