Answer
(a) $tan(x+y)=\frac{tanx+tany}{1-tanxtany}$
(b)$tan(x-y)=\frac{tanx-tany}{1+tanxtany}$
Work Step by Step
(a) $tan(x+y)=\frac{sin(x+y)}{cos(x+y)}$
$sin(x+y)=sinxcosy+cosxsiny$
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) $tan(x-y)=\frac{sin(x-y)}{cos(x-y)}$
$sin(x-y)=sinxcosy-cosxsiny$
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, $tan(x-y)=\frac{tanx-tany}{1+tanxtany}$
