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

Question

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?

Answer 1

$$\tan \left(x - \frac{\pi}{4}\right) = - \tan \left(x + \frac{\pi}{2}\right)$$

$$\tan \left(x - \frac{\pi}{4}\right) = \tan \left(-x - \frac{\pi}{2}\right)$$

$$x-\frac{\pi}{4}=-x-\frac{\pi}{2}+k\pi,\quad(k\in Z)$$

$$2x=-\frac{\pi}{4}+k\pi$$

$$x=-\frac{\pi}{8}+\frac{k\pi}{2}$$

Valid for any $k\in Z$.

Answer 2

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 $x=\pi n +\frac{3\pi}{8}$

Answer 3

Hint: Use that

$$\tan(x)+\tan(y)=\sec(x)\sec(y)\sin(x+y)$$

Answer 4

Hint

$$\tan x+\tan y=\frac{\sin(x+y)}{\cos x\cos y}$$