Question

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

• 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 $x\sqrt{1+x^2}$

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