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?
Best Answer
$$\tan \left(x - \frac{\pi}{4}\right) = - \tan \left(x + \frac{\pi}{2}\right)$$
$$\tan \left(x - \frac{\pi}{4}\right) = \tan \left(-x - \frac{\pi}{2}\right)$$
$$x-\frac{\pi}{4}=-x-\frac{\pi}{2}+k\pi,\quad(k\in Z)$$
$$2x=-\frac{\pi}{4}+k\pi$$
$$x=-\frac{\pi}{8}+\frac{k\pi}{2}$$
Valid for any $k\in Z$.