Question

The value of $\sin \left[\cot^{-1}\\{{\cos(tan^{-1}\,x)\\}}\right]$ is

• A. $\left(\sqrt{\frac{1+x^2}{2+x^2}}\right)$
• B. $\left(\sqrt{\frac{2+x^2}{1+x^2}}\right)$
• C. $\left(\sqrt{\frac{x^2-2}{x^2+1}}\right)$
• D. $\left(\sqrt{\frac{x^2-1}{x^2-2}}\right)$

Answer 1

Answer: (A)

$\left(\sqrt{\frac{1+x^2}{2+x^2}}\right)$

Answer 2

$cos\left(tan^{-1}x\right) = cos\,\theta$ if $x = tan \,\theta = \frac{1}{\sqrt{1+x^{2}}}$

$cot^{-1} \left[ cos\left(tan^{-1}x\right)\right] = cot^{-1}\left[\frac{1}{\sqrt{1+x^{2}}}\right] = \phi$
$\Rightarrow cot\,\phi = \frac{1}{\sqrt{1+x^{2}}}$

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