Trigonometry identities sum and substraction of cosine Someone can help me showing this identity?
$\frac{cos(a-b)}{cos(a+b)}=\frac{1+tan(a)tan(b)}{1-tan(a)tan(b)}$

We start with $\frac{1+tan(a)tan(b)}{1-tan(a)tan(b)}$:
$\frac{1+tan(a)tan(b)}{1-tan(a)tan(b)}=\frac{1+\frac{sin (a)sin (b)}{cos (a)cos (b)}}{1-\frac{sin (a)sin (b)}{cos (a)cos (b)}}=\frac{\frac{cos (a)cos (b)+sin (a)sin (b)}{cos (a)cos (b)}}{\frac{cos (a)cos (b)-sin (a)sin (b)}{cos (a)cos (b)}}=\frac{cos (a)cos (b)+sin (a)sin (b)}{cos (a)cos (b)-sin (a)sin (b)}=\frac{cos(a-b)}{cos(a+b)}$
