<s>[INST] Proving $\tan \left(\frac{\pi }{4} - x\right) = \frac{{1 - \sin 2x}}{{\cos 2x}}$ How do I prove the identity:
$$\tan \left(\frac{\pi }{4} - x\right) = \frac{{1 - \sin 2x}}{{\cos 2x}}$$
Any common strategies on solving other identities would also be appreciated.
I chose to expand the left hand side of the equation and got stuck here:
$$\frac{\cos x-\sin x}{\cos x+\sin x}$$
 [/INST] $$
\begin{aligned}
& \frac{1-\sin 2 x}{\cos 2 x} \\
=& \frac{(\cos x-\sin x)^{2}}{\cos ^{2} x-\sin ^{2} x} \\
=&\frac{(\cos x-\sin x)^{2}}{(\cos x+\sin x)(\cos x-\sin x)} \\
=& \frac{\cos x-\sin x}{\cos x+\sin x} \\
=& \frac{1-\tan x}{1+\tan x} \\
=& \tan \left(\frac{\pi}{4}-x\right)
\end{aligned}
$$
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