Question

If $0<x<1$, then $\sqrt{1+x^2}\left[\right\{xcos\left({\cot }^{-1}x\right)+\sin \left({\cot }^{-1}x\right){\}}^2-1{]}^{1/2}$ 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

The correct option is C $x\sqrt{1+x^2}$

$sin\left(co{t}^{-1}\right(x\left)\right)$
$=sin\left(si{n}^{-1}\frac{1}{\sqrt{1+x^2}}\right)$
$=\frac{1}{\sqrt{1+x^2}}$
and,
$=cos\left(co{t}^{-1}\right(x\left)\right)$
$=cos\left(co{s}^{-1}\right(\frac{x}{\sqrt{1+x^2}})$
$=\frac{x}{\sqrt{1+x^2}}$
Therefore,
Simplifying the above expression we get
$\sqrt{1+x^2}\left[\right(\frac{x^2+1}{\sqrt{1+x^2}}{)}^2-1{]}^{\frac{1}{2}}$

$=\sqrt{1+x^2}[x^2+1-1{]}^{\frac{1}{2}}$

$=\sqrt{1+x^2}.x$