Question

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

Answer: (C)

$\sqrt{1+x^2}{\left[{\left(x\cos {\cot }^{-1}x+\sin {\cot }^{-1}x\right)}^2-1\right]}^{1/2}$
$=\sqrt{1+x^2}{\left[{\left(x\cos {\cos }^{-1}\frac{x}{\sqrt{1+x^2}}+\sin {\sin }^{-1}\frac{1}{\sqrt{1+x^2}}\right)}^2-1\right]}^{1/2}$
$=\sqrt{1+x^2}{\left[{\left(\frac{x^2}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}\right)}^2-1\right]}^{1/2}$
$=\sqrt{1+x^2}{\left(x^2+1-1\right)}^{1/2}=x\sqrt{1+x^2}$