Question

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

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sec x+\tan x}$
• D. $\frac{1}{{\sec }^{2}x}$
• E. $\frac{1}{\tan x}$

Answer 1

The correct option is B $\frac{1}{2}$

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

$\Rightarrow y={\tan }^{-1}\left[\tan (\frac{\pi }{4}+\frac{x}{2})\right]$

$[\because \tan (\frac{\pi }{4}+\frac{x}{2})=\sec x+\tan x]$

$\Rightarrow y=\frac{\pi }{4}+\frac{x}{2}$

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

$\frac{dy}{dx}=0+\frac{1}{2}=\frac{1}{2}$