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

Using the angle sum formula for $\cos$:

$$\frac{\cos(a-b)}{\cos(a+b)}= \frac{\cos a \cos b + \sin a \sin b}{\cos a \cos b - \sin a \sin b}$$

Hint: Now all you need to do is to divide the numerator and denominator by $(\cos a\cos b)...$