Question:
Prove that:
$\frac{1-\cos 2 x+\sin 2 x}{1+\cos 2 x+\sin 2 \mathrm{x}}=\tan x$
Solution:
LHS $=\frac{1-\cos 2 x+\sin 2 x}{1+\cos 2 x+\sin 2 x}$
$=\frac{2 \sin ^{2} x+\sin 2 x}{2 \cos ^{2} x+\sin 2 x} \quad\left[\because 2 \sin ^{2} x=1-\cos 2 x\right.$ and $\left.2 \cos ^{2} x=1+\cos 2 x\right]$
$=\frac{2 \sin ^{2} x+2 \sin x \cos x}{2 \cos ^{2} x+2 \sin x \cos x} \quad(\because \sin 2 x=2 \sin x \cos x)$
$=\frac{2 \sin x(\sin x+\cos x)}{2 \cos x(\cos x+\sin x)}$
$=\tan \theta=\mathrm{RHS}$
Hence proved.
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