Question

If the derivative of $(x+\sec x)(x-\tan x)$ is $A(x+\sec x)+B(x-\tan x), \quad$ where $A$ and $B$ respectively are

• A. $\sec ^2 x-1,1+\sec x \tan x$
• B. $1-\sec ^2 x, 1-\sec x \tan x$
• C. $1-\sec ^2 x, 1+\sec x \tan x$
• D. $\sec ^2 x-1,1-\sec x \tan x$

Answer 1

Answer: (C)

Let $y=(x+\sec x)(x-\tan x)$
Differentiating $y$ w.r.t. $x$, we get
$\frac{d y}{d x} =(x+\sec x) \frac{d}{d x}(x-\tan x)+(x-\tan x) \frac{d}{d x}(x+\sec x) \text { (by product rule) } $
$ =(x+\sec x)\left(1-\sec ^2 x\right)+(x-\tan x)(1+\sec x \tan x)$