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

Question

Aicharya · Accepted Answer

Given, $\sqrt{1+x^2}\left[\left\{x \cos \left(\cot ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right\}^2-1\right]^{\frac{1}{2}}$
Let $\cot ^{-1} x=\theta$ $\begin{aligned} \therefore \quad & x=\cot \theta \\ & \Rightarrow \sin \theta=\frac{1}{\sqrt{1+x^2}}...(i) \\ & \Rightarrow \cos \theta=\frac{x}{\sqrt{1+x^2}} ...(ii)\end{aligned}$ $\begin{aligned} \therefore \quad & \sqrt{1+x^2}\left[\left\{x \cos \left(\cot ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right\}^2-1\right]^{\frac{1}{2}} \\ & =\sqrt{1+x^2}\left[\{x \cos \theta+\sin \theta\}^2-1\right]^{\frac{1}{2}} \\ & =\sqrt{1+x^2}\left[\left\{x \times \frac{x}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}\right\}^2-1\right]^{\frac{1}{2}} \\ & =\sqrt{1+x^2}\left[\left(\frac{x^2+1}{\sqrt{1+x^2}}\right)^2-1\right]^{\frac{1}{2}} \\ = & \sqrt{1+x^2}\left[\left(\sqrt{1+x^2}\right)^2-1\right]^{\frac{1}{2}} \\ = & \sqrt{1+x^2}\left(1+x^2-1\right)^{\frac{1}{2}} \\ = & x \sqrt{1+x^2}\end{aligned}$

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

Solution