Which of the following are the critical points $c$ of the function $y = \tan^{-1}(\sec x)$ in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$?

Only $c = 0$

The correct answer is Option (1) → Only $c = 0$ ##

$y = \tan^{-1}(\sec x)$

$y' = \frac{d}{dx}(\tan^{-1}(\sec x))$

$= \frac{1}{1 + \sec^2 x} \cdot \frac{d}{dx}(\sec x)$

$y' = \frac{\sec x \tan x}{1 + \sec^2 x}$

For critical points, put $y' = 0$,

$\frac{\sec x \tan x}{1 + \sec^2 x} = 0 \Rightarrow \sec x \tan x = 0$

Since $\sec x = \frac{1}{\cos x}$ is never zero, the equation reduces to:

$\tan x = 0 \Rightarrow x = 0$

At $x = 0$, $\sec 0 = 1$ and $\tan 0 = 0$, so $y' = 0$.

Thus, $x = 0$ is a critical point.

The derivative $y' = \frac{\sec x \tan x}{1 + \sec^2 x}$ is undefined only where $\sec x$ is undefined, i.e., at $x = \pm \frac{\pi}{2}$. However, these points are not included in the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Hence, the only critical point in $(-\frac{\pi}{2}, \frac{\pi}{2})$ is at $x = 0$.