If tanA = -4/3 and cosB = 15/17, where A lies in the 2nd quadrant, and B lies in the 4th quadrant.

Find the exact value of tan(A+B)

You know how to expand the tan(A+B) using the compound angles formula? I suggest you do so.

Then, to get tanB, you'll have to make a quick sketch of a right angled triangle, with angle B, adjacent side 15 and hypotenuse 17.

Work out the opposite side using Pythagoras' Theorem.

Then, find tan B which is opp/adj

Replace the values of tanA and tanB in your compound angle formula.

Post what you get! :)

Yea I did that, I was like "ARGGHHH HOW THE HELL DO YOU GET TANB???" I discovered shortly after I posted...

I ended up with 36/77... Idk if that's right.

*-36/77

Okay, I get:

\tan B = \frac{8}{15}

And since B is in the 4th quadrant, tan B is negative, hence:

\tan B = -\frac{8}{15}

Then;

\tan(A+B)= \frac{\tan A + \tan B}{1 - \tan A\tan B} = \frac{-\frac43 - \frac{8}{15}}{1 - \(-\frac43\)\(-\frac{8}{15}\)} = -\frac{84}{13}

You can confirm by using your calculator.

\tan^{-1}\(-\frac43\) = -53.1^o

which is 126.9 in the 2nd quadrant

\tan^{-1}\(-\frac{8}{15}\) = -28^o

which is 331.9 degrees in the 4th quadrant

Total, gives tan(126.9 + 331.9) = -6.45 which is about -84/13