Question

$\frac{d}{dx}\left(\frac{\cot x+\tan x}{\cot x-\tan x}\right)=$

• A. $2\sin 2x$
• B. $\sec 2x$
• C. $2\sec 2x\tan 2x$
• D. $2\cos 2x$

Answer 1

The correct option is C $2\sec 2x\tan 2x$

$\frac{cotx+tanx}{cotx-tanx}=\frac{\frac{cosx}{sinx}+\frac{sinx}{cosx}}{\frac{cosx}{sinx}-\frac{sinx}{cosx}}$

$=\frac{\frac{1}{sinxcosx}}{\frac{cos2x}{sinxcosx}}$

$=sec2x$

now $\frac{d}{dx}sec2x=2sec2xtan2x$