Question

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

• A. $\frac{x}{\sqrt{1+x^{2}}}$
• B. $x$
• C. $x \sqrt{1+x^{2}}$
• D. $\sqrt{1+ x ^{2}}$

Answer 1

Answer: (C)

$\sqrt{1+x^{2}}\left[\left(x \cos \cot ^{-1} x+\sin \cot ^{-1} x\right)^{2}-1\right]^{1 / 2}$
$=\sqrt{1+x^{2}}\left[\left(x \cos \cos ^{-1} \frac{x}{\sqrt{1+x^{2}}}+\sin \sin ^{-1} \frac{1}{\sqrt{1+x^{2}}}\right)^{2}-1\right]^{1 / 2}$
$=\sqrt{1+x^{2}}\left[\left(\frac{x^{2}}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+x^{2}}}\right)^{2}-1\right]^{1 / 2}$
$=\sqrt{1+x^{2}}\left(x^{2}+1-1\right)^{1 / 2}=x \sqrt{1+x^{2}}$