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

1. $\frac{x}{\sqrt{1+x^2}}$
2. $x$
3. $x \sqrt{1+x^2}$
4. $\sqrt{1+x^2}$

Solution

We have, $0 < x < 1$ Let $\quad \cot ^{-1} x=\theta$ $\Rightarrow \quad \cot \theta=x$ $\therefore \quad \sin \theta=\frac{1}{\sqrt{1+x^2}}=\sin \left(\cot ^{-1} x\right)$ and $\quad \cos \theta=\frac{x}{\sqrt{1+x^2}}=\cos \left(\cot ^{-1} x\right)$ Now, $\sqrt{1+x^2}\left[\left\{x \cos \left(\cot ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right\}^2-1\right]^{1 / 2}$
$ \text { Now, } \begin{aligned} & \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} \\ &=\sqrt{1+x^2}\left[\left\{x \frac{x}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}\right\}^2-1\right]^{1 / 2} \\ &=\sqrt{1+x^2}\left[\left(\frac{1+x^2}{\sqrt{1+x^2}}\right)^2-1\right]^{1 / 2} \\ &=\sqrt{1+x^2}\left[1+x^2-1\right]^{1 / 2} \\ &=x \sqrt{1+x^2} \end{aligned} $

Asked in: JEE Advanced 2008 (Paper 1)

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