Question

If $y={\tan }^{-1}x+{\cot }^{-1}x+{\sec }^{-1}x+{\csc }^{-1}x$, then $\frac{dy}{dx}$ is equal to

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

Answer 1

The correct option is C

$0$

Explanation for the correct option.

Find the value of $\frac{dy}{dx}$.

The equation can be simplified $y={\tan }^{-1}x+{\cot }^{-1}x+{\sec }^{-1}x+{\csc }^{-1}x$ using the relation ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\mathrm{\pi }}{2}$ and ${\sec }^{-1}x+{\csc }^{-1}x=\frac{\mathrm{\pi }}{2}$.

$\begin{array}{rcl}y& =& {\tan }^{-1}x+{\cot }^{-1}x+{\sec }^{-1}x+{\csc }^{-1}x\\ & =& \frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2}\\ & =& \mathrm{\pi }\end{array}$

Now differentiate both sides of the equation $y=\mathrm{\pi }$ with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left(\mathrm{\pi }\right)\\ & =& 0\end{array}$

Hence, the correct option is C.