# Thread: [SOLVED] tangent sum identity issue

1. ## [SOLVED] tangent sum identity issue

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 4. Thus tan(A) = $\displaystyle \frac{-3}{4}$.

for B, cosB=$\displaystyle \frac{-1}{4}$ so the hypotenuse is 4, the adjacent side -1, making the opposite side $\displaystyle \sqrt15$. Thus tan(b) = $\displaystyle \frac{\sqrt15}{-1}$ or just $\displaystyle -\sqrt15$.

The tangent sum identity is $\displaystyle \frac{tanA+tanB}{1-tanAtanB}$

Substituting in, I get
$\displaystyle \frac{\frac{-3}{4}+ -\sqrt15}{1-\frac{-3}{4}(-\sqrt15)}$

which is
$\displaystyle \frac{\frac{-3-4\sqrt15}{4}}{1-\frac{3\sqrt15}{4}}$
and then
$\displaystyle \frac{\frac{-3-4\sqrt15}{4}}{\frac{4-3\sqrt15}{4}}$
followed by
$\displaystyle \frac{-3-4\sqrt15}{4}*\frac{4}{4-3\sqrt15}$
and finally
$\displaystyle \frac{-3-4\sqrt15}{4-3\sqrt15}$

unfortunately, the answer would appear to be
$\displaystyle \frac{3-4\sqrt15}{4+3\sqrt15}$

Can someone please point out my mistake?

2. 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) = $\displaystyle \textcolor{red}{\frac{-3}{-4}}$.