Show that \(\frac{\sin 2x}{1 + \cos x}\) + \(\frac{sin2 x}{1 - cos x}\) is
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Correct Answer: Option C
Explanation:
\(\frac{\sin^{2} x}{1 + \cos x} + \frac{\sin^{2} x}{1 - \cos x}\)
\(\frac{\sin^{2} x (1 -Â \cos x) + \sin^{2} x (1 + \cos x)}{1 - \cos^{2} x}\)
= \(\frac{\sin^{2} x - \cos x \sin^{2} x + \sin^{2} x + \sin^{2} x \cos x}{\sin^{2} x}\)
(Note: \(\sin^{2} x + \cos^{2} x = 1\)).
= \(\frac{2 \sin^{2} x}{\sin^{2} x}\)
= 2.
\(\frac{\sin^{2} x}{1 + \cos x} + \frac{\sin^{2} x}{1 - \cos x}\)
\(\frac{\sin^{2} x (1 -Â \cos x) + \sin^{2} x (1 + \cos x)}{1 - \cos^{2} x}\)
= \(\frac{\sin^{2} x - \cos x \sin^{2} x + \sin^{2} x + \sin^{2} x \cos x}{\sin^{2} x}\)
(Note: \(\sin^{2} x + \cos^{2} x = 1\)).
= \(\frac{2 \sin^{2} x}{\sin^{2} x}\)
= 2.