Solve $\tan {(x-\frac{\pi}{4}})=-\tan{(x+\frac{\pi}{2})}$ The question:

Without the use of a calculator, solve for all values of $x$ if $\tan {(x-\frac{\pi}{4}})=-\tan{(x+\frac{\pi}{2})}.$

Using the compound angle formula for solving equations is normally easy, but I stumbled across this problem. 
The $LHS$ is easy to expand, but when you apply the compound formula for the $RHS$,
\begin{align}
\tan{(x+\frac{\pi}{2})} & = \frac{\tan{(x)} + \tan{(\frac{\pi}{2})}}{1-\tan{(x)}\cdot\tan{(\frac{\pi}{2})}} \\
\end{align}
You might notice that this is a problem because I cannot evaluate $\tan{(\frac{\pi}{2})}$. So this is what I tried. First I tried writing 
\begin{align}
\tan{(x+\frac{\pi}{2})} & = \frac{\sin{(x+\frac{\pi}{2})}}{\cos{(x+\frac{\pi}{2})}} \\
& = \frac{\cos (x)}{\sin (x)}
\end{align}
which I knew was wrong. Anyone know how to get around this?

Hint: $\tan(x)$ is an odd function so $\tan(-x)=-\tan(x)$
Solution:

 By above we get $\tan{(x-\pi/4)}=\tan{(-x-\pi/2)}$ so
 $$x-\pi/4 \equiv -x-\pi/2 \pmod{\pi}$$
 (since the tangent function has a period of $\pi$)
 $$2x \equiv -\frac{\pi}{4} \equiv \frac{3\pi}{4} \equiv \frac{7\pi}{4}\pmod{\pi}$$
 Solving this yields solutions $x=\pi n +\frac{7\pi}{8}$ and $x=\pi n +\frac{3\pi}{8}$

