Question

The derivative of $\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) $ with respect to $\cot^{-1} \left(\frac{1-3x^{2}}{3x-x^{3}}\right)$ is

• A. $ \frac{1}{2}$
• B. $1$
• C. $ \frac{-1}{2}$
• D. $ - \frac{2}{3}$

Answer 1

The Correct Option is D

Let $y = \cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$ ...(i)
and $ z = \cot^{-1} \left(\frac{1-3x^{2}}{3x-x^{3}}\right) $ ....(ii)
Putting $x = \tan \theta$ in (i) and $x = \cot \, t$ in (ii), respectively
$y = \cos ^{-1}\left(\frac{1-\tan^{2} \theta}{1+\tan ^{2} \theta }\right) $
$= \cos ^{-1} \left(\cos 2\theta\right) =2 \tan^{-1} x$
$ z = \cot^{-1}\left(\frac{1-3 \cot ^{2} t }{3 \cot t -\cot^{3} t }\right) $
$=\cot ^{-1} \left(\cot 3t\right) =3 \cot ^{-1} x$
$\frac{dy}{dx}= \frac{2}{1+x^{2}} $ and $\frac{dz}{dx} = \frac{-3}{1+x^{2}}$
$ \frac{dy}{dz} = \frac{dy}{dx}. \frac{dx}{dz} = \left(\frac{2}{1+x^{2}}\right)\left(\frac{1+x^{2}}{-3}\right)= \frac{-2}{3}$