If 0 < x < 1, then $\sqrt{1+x^2} \left[\begin{Bmatrix}x cos (cot^{-1}x)+sin(cot^{-1}x\end{Bmatrix}^2-1\right]^{1/2}$
Answer & explanation
Correct answer: option 3
$ x cos (cot^{-1}x) + sin (cot^{-1}x)$
$= x cos \left(cos^{-1}\frac{x}{\sqrt{1+x^2}}\right) + sin \left( sin^{-1}\frac{1}{\sqrt{1+x^2}}\right)$
$=\frac{x^2}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}= \sqrt{x^2+1}$
$∴\sqrt{1+x^2} \left[\begin{Bmatrix}x cos (cot^{-1}x)+sin(cot^{-1}x\end{Bmatrix}^2-1\right]^{1/2}$
$= \sqrt{1+x^2} (x^2 + 1 - 1)^{1/2}= x\sqrt{1+x^2}$