Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

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