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