Question

$\frac{d}{dx}\left[\frac{tanx-cotx}{tanx+cotx}\right]=$

• A. 2 sin2x
• B. -2 sin2x
• C. 2 cos 2x
• D. -2cos 2x

Answer 1

The correct option is A 2 sin2x

$\frac{d}{dx}\left(\frac{\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}}{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}\right)=\frac{d}{dx}\left(\frac{{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x+{\cos }^{2}x}\right)$

$=\frac{d}{dx}({\sin }^{2}x-{\cos }^{2}x)=2\sin x\cos x+2\cos x\sin x=2\sin 2x\Rightarrow \left(A\right)$