Show that \(\frac{\sin 2x}{1 + \cos x}\) + \(\frac{sin2 x}{1 - cos x}\) is

A.

sin x

B.

cos2x

C.

2

D.

3

Correct answer is C

\(\frac{\sin^{2} x}{1 + \cos x} + \frac{\sin^{2} x}{1 - \cos x}\)

\(\frac{\sin^{2} x (1 - \cos x) + \sin^{2} x (1 + \cos x)}{1 - \cos^{2} x}\)

= \(\frac{\sin^{2} x - \cos x \sin^{2} x + \sin^{2} x + \sin^{2} x \cos x}{\sin^{2} x}\)

(Note: \(\sin^{2} x + \cos^{2} x = 1\)).

= \(\frac{2 \sin^{2} x}{\sin^{2} x}\)

= 2.