Q. If $0 < x < 1$, then $\sqrt{1 + x^{2}}\left\lbrack \left\{ x\cos{\left( \cot^{- 1}x \right) + \sin\left( \cot^{- 1}x \right)} \right\}^{2} - 1 \right\rbrack^{1/2}$ is equal to
A) $\frac{x}{\sqrt{1 + x^{2}}}$
B) $x$
C) $x\sqrt{1 + x^{2}}$
D) $\sqrt{1 + x^{2}}$
Explanation
(c)
$\sqrt{1 + x^{2}}\left\lbrack \left\{ x\cos{\left( \cot^{- 1}x \right) + \sin\left( \cot^{- 1}x \right)} \right\}^{2} - 1 \right\rbrack^{1/2}$
$= \sqrt{1 + x^{2}}\left\lbrack \left\{ x\cos{\left( \cos^{- 1}\frac{x}{\sqrt{1 + x^{2}}} \right) + \sin\left( \sin^{- 1}\frac{1}{\sqrt{1 + x^{2}}} \right)} \right\}^{2} - 1 \right\rbrack^{1/2}$
$= \sqrt{1 + x^{2}}\left\lbrack \left\{ x.\frac{x}{\sqrt{1 + x^{2}}} + \frac{1}{\sqrt{1 + x^{2}}} \right\}^{2} - 1 \right\rbrack^{1/2}$
$= \sqrt{1 + x^{2}}\left\lbrack 1 + x^{2} - 1 \right\rbrack^{1/2}$
$= x\sqrt{1 + x^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.