If $\tan A=-\frac 1 2$ and $\tan B =-\frac 1 3$, then $A+B =?$

Question

Answer should be in radians Like π/4 (45°) π(90°). I used $\tan(A+B)$ formula and got $5/7$ as the answer, but that's obviously wrong.

Answer 1

Using the $\tan(A+B)$ formula,

$$ \tan(A+B) = \frac{-1/2 - 1/3}{1-(1/2)(1/3)} = -1 $$

now use the $\arctan$,

$$ A+B = \arctan(-1) = n\pi + \frac{3\pi}{4}, \ \ n\in\mathbb Z $$

Answer 2

use that $$\tan(A+B)=\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$$

Answer 3

Use the formula $\tan (x+y)=\frac{(\tan x+ \tan y)}{1-\tan x\tan y}$

And then take inverse

Answer 4

Another approach:

$\tan a + \tan b = \frac{\sin(a+b)}{\cos a \cos b}$

Then

$\tan a + \tan b = \frac{\sin(a+b)}{\cos a \cos b}$

$\cos (\tan^{-1}x) = \frac{1}{\sqrt{x^2+1}}$

$-\frac{1}{2} + -\frac{1}{3} = \frac{\sin(a+b)}{\frac{1}{\sqrt{(\frac{-1}{2})^2+1}}\cdot \frac{1}{\sqrt{(\frac{-1}{3})^2+1}}}$

$-5 = \frac{\sin(a+b)}{\frac{1}{\sqrt{5}}\cdot \frac{1}{\sqrt{10}}}$

$-\frac{1}{\sqrt{2}} = \sin(a+b)$

$\sin^{-1}\frac{-\sqrt2}{2}= \frac{-\pi}{4}$