Question

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

• A. $-1$
• B. $\pi$
• C. $0$
• D. $1$

Answer 1

Answer: (C)

$0$

Consider the given equation.

$y={\tan }^{-1}x+{\cot }^{-1}x+{\sec }^{-1}x+{\csc }^{-1}x$

On differentiating w.r.t $x$, we get

$\frac{dy}{dx}=\frac{1}{1+x^2}-\frac{1}{1+x^2}+\frac{1}{\left|x\right|\sqrt{x^2-1}}-\frac{1}{\left|x\right|\sqrt{x^2-1}}$

$\frac{dy}{dx}=0$

Hence, this is the answer.