Question

Prove that:
$\frac{1-\cos 2x+\sin 2x}{1+\cos 2x+\sin 2\mathrm{x}}=\tan x$

Answer 1

$\mathrm{LHS}=\frac{1-\cos 2x+\sin 2x}{1+\cos 2x+\sin 2x}$

$$\begin{gathered} =\frac{2{\sin }^{2}x+\sin 2x}{2{\cos }^{2}x+\sin 2x}\left[\because 2{\sin }^{2}x=1-\cos 2x\mathrm{and}2{\cos }^{2}x=1+\cos 2x\right] \\ =\frac{2{\sin }^{2}x+2\sin x\cos x}{2{\cos }^{2}x+2\sin x\cos x}\left(\because \sin 2x=2\sin x\cos x\right) \\ =\frac{2\sin x\left(\sin x+\cos x\right)}{2\cos x\left(\cos x+\sin x\right)} \\ =\mathrm{tan\theta }=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$