Question

If $y=\tan ^{-1} x+\cot ^{-1} x+\sec ^{-1} x+\operatorname{cosec}^{-1} x$, then $\frac{d y}{d x}$ is equal to

• A. $\frac{x^2-1}{x^2+1}$
• B. $\pi$
• C. 0
• D. 1

Answer 1

Answer: (C)

$y=\tan ^{-1} x+\cot ^{-1} x+\sec ^{-1} x+\operatorname{cosec}^{-1} x$
$=\frac{\pi}{2}+\frac{\pi}{2} \left(\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right)$
$\left(\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2}\right)$
$\therefore y=\pi \Rightarrow \frac{d y}{d x}=0$