## Calculus 8th Edition

(a) $tan(x+y)=\frac{tanx+tany}{1-tanxtany}$ (b) $tan(x-y)=\frac{tanx-tany}{1+tanxtany}$
(a) $tan(x+y)=\frac{sin(x+y)}{cos(x+y)}$ Since, $sin(x+y)=sinxcosy+cosxsiny$ and $cos(x+y)=cosxcosy-sinxsiny$ Thus, $tan(x+y)=\frac{sinxcosy+cosxsiny}{cosxcosy-sinxsiny}$ Divide the numerator and denominator by $cosx$ $cosy$: $tan(x+y)=\frac{\frac{sinx}{cosx}+\frac{siny}{cosy}}{1-\frac{sinx siny}{cosxcosy}}$ Hence, $tan(x+y)=\frac{tanx+tany}{1-tanxtany}$ (b) $tan(x-y)=\frac{sin(x-y)}{cos(x-y)}$ Since, $sin(x-y)=sinxcosy-cosxsiny$ and $cos(x-y)=cosxcosy+sinxsiny$ Thus, $tan(x-y)=\frac{sinxcosy-cosxsiny}{cosxcosy+sinxsiny}$ Divide the numerator and denominator by $cosx$ $cosy$: $tan(x-y)=\frac{\frac{sinx}{cosx}-\frac{siny}{cosy}}{1+\frac{sinx siny}{cosxcosy}}$ Hence, $tan(x+y)=\frac{tanx-tany}{1+tanxtany}$