Question 14061

Aug 4, 2016

No. Your suggestion is close, but the $\cos A \cos B$ term should not be dividing the 1.

$\tan \left(A + B\right) = \frac{\tan \left(A\right) + \tan \left(B\right)}{1 - \tan \left(A\right) \tan \left(B\right)}$

Explanation:

Let's derive it. We know that

$\sin \left(A + B\right) = \sin \left(A\right) \cos \left(B\right) + \sin \left(B\right) \cos \left(A\right)$

$\cos \left(A + B\right) = \cos \left(A\right) \cos \left(B\right) - \sin \left(A\right) \sin \left(B\right)$

$\tan \left(A + B\right) = \frac{\sin \left(A + B\right)}{\cos \left(A + B\right)}$

$= \frac{\sin \left(A\right) \cos \left(B\right) + \sin \left(B\right) \cos \left(A\right)}{\cos \left(A\right) \cos \left(B\right) - \sin \left(A\right) \sin \left(B\right)}$

Take $\cos \left(A\right) \cos \left(B\right)$ out of denominator as common factor:

$= \frac{\sin \left(A\right) \cos \left(B\right) + \sin \left(B\right) \cos \left(A\right)}{\cos \left(A\right) \cos \left(B\right) \left(1 - \tan \left(A\right) \tan \left(B\right)\right)}$

We can split this up into two separate terms by splitting up the numerator. This means we can cancel out the cosines on the numerator.

=(sin(A)cancel(cos(B)))/(cos(A)cancel(cos(B))(1 - tan(A)tan(B))

 + (sin(B)cancel(cos(A)))/(cancel(cos(A))cos(B)(1 - tan(A)tan(B))#

Because $\tan \phi = \frac{\sin \phi}{\cos \phi}$ this leaves:

$\frac{\tan \left(A\right)}{1 - \tan \left(A\right) \tan \left(B\right)} + \frac{\tan \left(B\right)}{1 - \tan \left(A\right) \tan \left(B\right)}$

$\tan \left(A + B\right) = \frac{\tan \left(A\right) + \tan \left(B\right)}{1 - \tan \left(A\right) \tan \left(B\right)}$