Question

Derivative of $\tan^{-1} \left(\frac{t}{1+z}\right) $ w.r.t. $\tan^{-1} \left(\frac{z}{1+t}\right) $, where $t = \sin x , z = \cos x$ is

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (A)

Put $u=\tan^{-1} \frac{t}{1+z}$ = $\tan^{-1} \frac{\sin x}{1+\cos x}$
$ =\tan^{-1}\tan \frac{x}{2} = \frac{x}{2} $
$\therefore \frac{du}{dx} = \frac{1}{2 }$
Let $ v =\tan^{-1} \frac{z}{1+t} =\tan^{-1} \frac{\cos x}{1+\sin x}$
$ =\tan^{-1} \frac{\cos^{2} \frac{x}{2} -\sin^{2} \frac{x}{2} }{\left(\cos \frac{x}{2} + \sin \frac{x}{2}\right)^{2}}$
$ =\tan^{-1} \frac{\cos \frac{x}{2} -\sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}}$
$ = \tan^{-1} \frac{1- \tan \frac{x}{2}}{1+ \tan \frac{x}{2}} =\tan^{-1} \tan \left(\frac{\pi}{4} - \frac{\pi}{2}\right) $
$= \frac{\pi}{4} - \frac{\pi}{2}$
$ \therefore \frac{dv}{dx}=- \frac{1}{2}$
$ \therefore \frac{dv}{dx} = \frac{du/dx}{dv/dx} = \frac{\frac{1}{2}}{-\frac{1}{2}} = - 1$