Originally Posted by

**satis** I've been working through a tanget sum identity problem and I've mostly got it, except I've got the signs reversed. I was curious if someone could point out the mistake I made. I apologize for how long this is. All assistance is appreciated.

Problem is

$\displaystyle tan(A+B)$

where

$\displaystyle sinA = \frac{-3}{5}$

is in QIII

and

$\displaystyle cosB = \frac{-1}{4}$

is in QII

sinA= $\displaystyle \frac{-3}{5}$ so the hypotenuse is 5, the oppose side is -3, making the adacent side (negative) 4. Thus tan(A) = $\displaystyle \textcolor{red}{\frac{-3}{-4}}$.

here is your mistake ... in quad III, tangent is positive

Can someone please point out my mistake?