Question

I need help never got taugh on hot to do this

Answer 1

Given the function

`\tan ((5\pi)/(6))`

We can find the corresponding value below;

Since

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

Answer:

`\tan (-(\pi)/(6))`