1. ## Trig question?

How do I solve this?
Given $tan(\pi-t)=10$ tind $tan t$

I have no clue where to start.

2. Originally Posted by dorkymichelle
How do I solve this?
Given $tan(\pi-t)=10$ tind $tan t$
I have no clue where to start.

$\tan(\pi-t)= \frac{\tan\pi - \tan t}{1+\tan\pi\tan t} =
\frac{0-\tan t}{1+0} = -\tan t$

Therefore, $\tan(\pi-t)=10 \implies \tan t = -10$.

Another way to tackle this is using arctan, but using properties has its advantages.

-Andy