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Jebto · Jan 26, 2011, 11:06 PM

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)

Unknown008 · Jan 26, 2011, 11:23 PM

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! :)

Jebto · Jan 26, 2011, 11:44 PM

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.

Jebto · Jan 26, 2011, 11:50 PM

*-36/77

Unknown008 · Jan 27, 2011, 12:22 AM

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