Prove that \[ \frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x}=\tan x \]
Proof:
\[
LHS=\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x}
\]
Using the identities:
\[
1-\cos2x=2\sin^2x
\]
\[
1+\cos2x=2\cos^2x
\]
\[
\sin2x=2\sin x\cos x
\]
Substituting these values:
\[
LHS=\frac{2\sin^2x+2\sin x\cos x}{2\cos^2x+2\sin x\cos x}
\]
Taking common factors:
\[
LHS=\frac{2\sin x(\sin x+\cos x)}{2\cos x(\cos x+\sin x)}
\]
Cancel common terms:
\[
LHS=\frac{\sin x}{\cos x}
\]
\[
=\tan x
\]
Hence proved,
\[
\boxed{\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x}=\tan x}
\]