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tan A = -1/2 , tan B = -1/3 , then A+B = ?

Trigonometry
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I had come up with : tan (A+B) = -1
so (A+B) = ?

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Other answers:

A+B = -45 + 180 * k
is it, 7pi /4
\[\tan(A+B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A) \cdot \tan(B)}\]
-45 = pi/4 180 = pi so -pi/4 + pi = 3pi/4 -pi/4 + 2pi = 7pi/4 and so on..
If you calculated it right then: \[A+B= \tan^{-1}(-1)\]
so 7pi/4 is one possibility
\[A+B = \tan^{-1}\frac{ -1 }{ 2 }+\tan^{-1} \frac{ -1 }{ 3 }\]
so it can be : 3 pi / 4, 7pi/4 , . ... It can not be 5 pi / 4 because it will be positive for tan.
But the book says that the answer is : d) none Options were : i) pi/4 ii) 3 pi/4 iii) 5 pi/4 iv) none.
So it will be 3 pi/4 ? or none?
it will be 3pi/4 are any constraints given?
constraints?
like A+B should be between 0 to 90 or something like this ?
else 3pi/4 is correct
ok thanks a lot @hartnn :)
wud tan(A+B) formula work for all angles
yes.
ok thank you :D
\[\pi - \frac{\pi}{4} = \frac{3 \pi}{4}\]
There was some mistake in the formula that I had written above.. That day I was looking for the same and now I have found it : \[\tan^{-1}(-x) = \pi - \tan^{-1}(x)\]

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