Question

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

• 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

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

Explanation for correct option:

Step-1: Assume ${\cot }^{-1}\left(x\right)=\theta$ and finding $\sin \left(\theta \right),\cos \left(\theta \right)$

Let, $$\begin{gathered} {\cot }^{-1}\left(x\right)=\theta \\ x=\cot \left(\theta \right) \end{gathered}$$

$\Rightarrow$$\sin \left(\theta \right)=\frac{1}{\sqrt{1+x^2}}........\left(i\right)$

$\Rightarrow$$\cos \left(\theta \right)=\frac{x}{\sqrt{1+x^2}}........\left(ii\right)$

Step-2: Simplify the given data.

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

$$\begin{gathered} =\sqrt{1+x^2}{\left[{\left\{x\cos \left({\cot }^{-1}\left(x\right)\right)+\sin \left({\cot }^{-1}\left(x\right)\right)\right\}}^2-1\right]}^{\frac{1}{2}} \\ =\sqrt{1+x^2}{\left[{\left\{x\times \frac{x}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}\right\}}^2-1\right]}^{\frac{1}{2}}\left(fromequation\left(i\right)\&\left(ii\right)\right) \\ =\sqrt{1+x^2}{\left[{\left\{\frac{x^2+1}{\sqrt{1+x^2}}\right\}}^2-1\right]}^{\frac{1}{2}} \\ =\sqrt{1+x^2}{\left[{\left\{\sqrt{1+x^2}\right\}}^2-1\right]}^{\frac{1}{2}} \\ =\sqrt{1+x^2}{\left[1+x^2-1\right]}^{\frac{1}{2}} \\ =x\sqrt{1+x^2} \end{gathered}$$

Hence, correct answer is option $\left(c\right)$