Question

I need help never got taugh on hot to do this

Answer 1

Given the function

$\tan (\frac{5\pi}{6})$

We can find the corresponding value below;

Since

$\begin{gathered} \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B} \\ we\text{ can have } \\ \tan (\frac{5\pi}{6})=\tan (\pi-\frac{\pi}{6}) \\ \therefore\tan (\pi-\frac{\pi}{6})=\frac{\tan \pi-\tan \frac{\pi}{6}}{1+\tan \pi\tan \frac{\pi}{6}} \\ \text{ since tan}\pi=0 \\ \therefore\tan (\pi-\frac{\pi}{6})=\frac{0-\tan \frac{\pi}{6}}{1+0\times\tan \frac{\pi}{6}} \\ \tan (\pi-\frac{\pi}{6})=0-\tan \frac{\pi}{6} \\ \tan (\pi-\frac{\pi}{6})=-\tan \frac{\pi}{6} \\ \tan (\pi-\frac{\pi}{6})=\tan (-\frac{\pi}{6}) \\ \therefore\tan (\frac{5\pi}{6})=\tan (\pi-\frac{\pi}{6})=\tan (-\frac{\pi}{6}) \end{gathered}$

Answer:

$\tan (-\frac{\pi}{6})$