Question

If $0<x<1$, then $\sqrt{1+x^2}{\left[{\left\{x\cos \left(co{t}^{-1}x\right)+\sin \left(co{t}^{-1}x\right)\right\}}^2-1\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=?$

• 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

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

Step 1. Find the value of $\sqrt{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}{\left[{\left\{xcos\left(co{t}^{-1}x\right)+sin\left(co{t}^{-1}x\right)\right\}}^2-1\right]}^{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}$:

Let $co{t}^{-1}x=\theta$

$\Rightarrow$ $cot\theta =x$

$\Rightarrow$ $\begin{array}{rcl}\sin \theta & =& \frac{1}{\sqrt{1+x^2}}\\ & =& \sin (co{t}^{-1}x)\end{array}$

and

$\begin{array}{rcl}\cos \theta & =& \frac{x}{\sqrt{1+x^2}}\\ & =& \cos (co{t}^{-1}x)\end{array}$

Step 2. Put the values of $\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\theta }$ and $\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }$ in given expression, we get

$\therefore \sqrt{1+x^2}{\left[{\left\{x\cos \left(co{t}^{-1}x\right)+\sin \left(co{t}^{-1}x\right)\right\}}^2-1\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$$\begin{gathered} =\sqrt{1+x^2}{\left[{\left(x\times \frac{x}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}\right)}^2–1\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =\sqrt{1+x^2}{\left[{\left(\frac{1+x^2}{\sqrt{1+x^2}}\right)}^2–1\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =\sqrt{1+x^2}{\left(1+x^2–1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =\mathit{x}\mathbf{}\sqrt{\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{x}}^{\mathbf{2}}} \end{gathered}$$

Hence, Option ‘C’ is Correct.