Differentiate the function $\tan^{-1}(\sec x + \tan x), \quad -\frac{\pi}{2} < x < \frac{\pi}{2}$ with respect to $x$.

$\frac{1}{2}$

The correct answer is Option (2) → $\frac{1}{2}$ ##

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

$∴\frac{dy}{dx} = \frac{d}{dx} \tan^{-1}(\sec x + \tan x)$

$= \frac{1}{1 + (\sec x + \tan x)^2} \cdot \frac{d}{dx} (\sec x + \tan x) \quad \left[ ∵\frac{d}{dx} (\tan^{-1} x) = \frac{1}{1 + x^2} \right]$

$= \frac{1}{1 + \sec^2 x + \tan^2 x + 2\sec x \cdot \tan x} \cdot [\sec x \cdot \tan x + \sec^2 x]$

$\left[ ∵\frac{d}{dx} \sec x = \sec x \cdot \tan x, \quad \frac{d}{dx} \tan x = \sec^2 x \right]$

$= \frac{1}{(\sec^2 x + \sec^2 x + 2\sec x \cdot \tan x)} \cdot \sec x \cdot (\sec x + \tan x) \quad [∵1 + \tan^2 x = \sec^2 x]$

$= \frac{1}{2 \sec x (\tan x + \sec x)} \cdot \sec x (\sec x + \tan x) = \frac{1}{2}$