Question

prove that tan(a + b) = tan(a)+tan(b)/ 1-tan(a)tan(b)

Answer 1

I think you have to use the sum identity for sin and cos\[\frac{\sin(a+b)}{\cos(a+b)}\]\[\frac{\sin(a) \cos(b) + \cos(a)\sin(b)}{\cos(a)\cos(b)-\sin(a)\sin(b)}\]

Then divide numerator and denominator by cos(a)sin(b)
\[\frac{\frac{\sin(a)\cos(b)+\cos(a)\sin(b)}{\cos(a)\cos(b)}}{....}\]
This give you a numerator of
\[\frac{\tan(a)+\tan(b)}{....}\]