Continuity and Differentiability — Class 12 Maths Solution

Let $y = {\tan ^{ - 1}}(\sec x + \tan x)$
therefore,$\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\tan ^{ - 1}}(\sec x + \tan x)$

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

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

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

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