Question

IF $y=ta{n}^{-1}(secx+tanx)$ then $\frac{dy}{dx}=$

• A. $1$
• B. $\frac{1}{2}$
• C. $-1$
• D. $0$

Answer 1

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

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

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

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

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

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

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

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

$=\frac{1}{2}$