Question

Prove that \[\tan(x+y) = \frac{\tan(x) + \tan(y)}{1 - \tan(x) \tan(y)}\]

Answer 1

use the addition formulas for sin and cos.

Answer 2

\[\frac{\sin(X+y)}{\cos(X+Y)}\]
=\[\frac{sixcosy+cosxsiny}{cosxcosy-sinxsiny}\]
dividing numerator and denominator by cosxcosy we get
\[\frac{tanx+tany}{1-tanxtany}\]