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

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Aicharya · Accepted Answer

Let $\cot ^{-1} x=	heta$, then $x=\cot 	heta$ and $\begin{aligned} & \frac{\pi}{4} \lt 	heta \lt \frac{\pi}{2} \quad \ldots[\because 0 \lt x \lt 1 \Rightarrow 0 \lt \cot 	heta \lt 1] \ 	herefore \quad & \sqrt{1+x^2}\left[\left\{x \cos \left(\cot ^{-1} xight)+\sin \left(\cot ^{-1} xight)ight\}^2-1ight]^{\frac{1}{2}} \ = & \sqrt{1+x^2}\left[\{\cot 	heta \cos 	heta+\sin 	heta\}^2-1ight]^{\frac{1}{2}} \ = & \sqrt{1+x^2} \cdot\left\{\left(\frac{\cos ^2 	heta+\sin ^2 	heta}{\sin 	heta}ight)^2-1ight]^{\frac{1}{2}} \ = & \sqrt{1+x^2} \sqrt{\operatorname{cosec}^2 	heta-1} \ = & \sqrt{1+x^2}|\cot 	heta| \ = & \left(\sqrt{1+x^2}ight)|x| \ = & x \sqrt{1+x^2} \end{aligned}$ $\ldots[\because 0 \lt x \lt 1]$

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

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