TITLE: The identity $\tan({\pi\over4}-{a\over2}) = \sec(a)-\tan(a)$
QUESTION [13 upvotes]: Today I was solving an exercise and while checking the solution on WolframAlpha, the website used the following identity:
$$\tan \left({\pi\over4}-{\alpha\over2} \right) = \sec(\alpha)-\tan(\alpha)$$
Since I've never seen that formula, I tried to calculate it with the identities I know, especially using
$$\tan \left({\pi\over4}-{\alpha\over2} \right) = {(1-\tan({\alpha\over2}))\over(1+\tan({\alpha\over2}))},$$
but I couldn't make it work.
Is this a known property of trigonometry, like for example $\cos({\pi\over2}-x)=\sin(x)$?

REPLY [6 votes]: Here's a couple of trigonographs:

$$\begin{align}
\sec\theta - \tan\theta \;=\; \tan\phi \;=\; \tan\frac{90^\circ - \theta}{2} \;=\; \tan\left( 45^\circ - \frac{\theta}{2}\;\right) \\[4pt]
\sec\theta + \tan\theta \;=\; \tan\psi \;=\; \tan\frac{90^\circ + \theta}{2} \;=\; \tan\left( 45^\circ + \frac{\theta}{2}\;\right)
\end{align}$$