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:
$$\frac{\cos x-\sin x}{\cos x+\sin x}$$
 A: Use the identity $$\tan(x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$$.  $$ \text{Multiply  the  expression} \ \frac{\cos x-\sin x}{\cos x+\sin x}\ \text{by} \ \frac{\cos x-\sin x}{\cos x-\sin x}.$$  You'll get an identity in the denominator immediately and finish the numerator.
A: As @ChristopherErnst suggests in a comment, some things become "more obvious after experience". Here are two alternative approaches to your problem that bear this out.

If you find yourself working with double- and half-angle arguments often, you might get the immediate sense that the right-hand side would be better if the roles of sine and cosine were reversed, since
$$\frac{1-\cos 2\theta}{\sin 2\theta} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta \qquad\qquad(*)$$
As it turns out, reversing those roles is as easy as replacing the arguments with their complements:
$$\cos x = \sin\left(\frac{\pi}{2}-x\right) \qquad \sin x = \cos\left(\frac{\pi}{2}-x\right)$$
So, writing $2x= \frac{\pi}{2}-2\theta$ (equivalently, $2\theta = \frac{\pi}{2}-2x$) turns your identity into $(*)$.

Another approach immediately recognizes "$1-\sin 2x$" as a perfect square:
$$1-\sin 2x = (\cos^2 x + \sin^2 x) - 2 \cos x \sin x = \left( \cos x - \sin x \right)^2$$
(This isn't something I've seen exploited all that often, but it has come up with unusual frequency in some trig manipulations in my current research, so it kinda jumps out at me.) It's a convenient counterpart to the difference-of-squares version of the cosine double-angle identity:
$$\cos 2x = \cos^2 x - \sin^2 x = \left( \cos x - \sin x \right)\left( \cos x + \sin x \right)$$
Thus, the right-hand side of your identity reduces nicely ...
$$\frac{\left( \cos x - \sin x \right)^2}{\left( \cos x - \sin x \right)\left( \cos x + \sin x \right)} = \frac{\cos x - \sin x}{\cos x + \sin x}$$
... to the expression you have already shown to be equal to $\tan\left(\frac{\pi}{4}-x\right)$.
A: $$
\begin{aligned}
& \frac{1-\sin 2 x}{\cos 2 x} \\
=& \frac{(\cos x-\sin x)^{2}}{\cos ^{2} x-\sin ^{2} x} \\
=&\frac{(\cos x-\sin x)^{2}}{(\cos x+\sin x)(\cos x-\sin x)} \\
=& \frac{\cos x-\sin x}{\cos x+\sin x} \\
=& \frac{1-\tan x}{1+\tan x} \\
=& \tan \left(\frac{\pi}{4}-x\right)
\end{aligned}
$$
