Question

Let sin A = 12/13 with 90º≤A≤180º and tan B = -4/3 with 270º≤B≤360º. Find tan (A + B).

Answer 1

I keep getting -0.06525 as my answer, which isn't an option. WHERE AM I GOING WRONG

Answer 2

This is the correct equation right? \[\tan(a+b) = \frac{ \tan a + \tan b }{ 1- \tan a*\tan b }\]

Answer 3

Tan A is between 90 and 180 so its actually

Answer 4

I so have this for you. Let me draw a couple of things then explain, ok?
The formula to find tan (A + B) is\[\tan(A + B) =\frac{ tanA+tanB }{ 1-tanAtanB }\]
The tangent of angle A is -12/5 and the tangent of angle B is -4/3. Fill in your formula so it looks like this:\[\frac{ (-12/5)+(-4/3) }{ 1-(-12/5)(-4/3) }\]
Now we will do some math here. Getting a common denominator for the numerator of 15 gives you
We will now go ahead and find the common denominator in the denominator of your rational expression.
Now we can solve both top and bottom of this mess, right? -36/15 + -20/15 = -56/15. And in the denominator 15/15 - 48/15 = -33/15.
Cancel out the 15's and you're left with a -56/-33, which means the negatives cancel out leaving a positive 56/33. Get that? It's the third answer down in your choices.