Question

Proove this identities
Sin(a+b) / cos(a-b) = tan a + tan b / 1 + tan a * tan b

Answer 1

Question

$$\begin{array}{l} \frac{\sin(a + b)}{\cos(a - b)} = \frac{\tan a + \tan b}{1 + \tan a \cdot \tan b} \\ \frac{\sin(a + b)}{\cos(a - b)} = \frac{\sin a \cdot \cos b + \cos a \cdot \sin b}{\cos a \cdot \cos b + \sin a \cdot \sin b} = \frac{\left(\sin a \cdot \cos b + \cos a \cdot \sin b\right) / \cos a \cdot \cos b}{\left(\cos a \cdot \cos b + \sin a \cdot \sin b\right) / \cos a \cdot \cos b} = \\ = \frac{\sin a \cdot \cos b / \cos a \cdot \cos b + \cos a \cdot \sin b / \cos a \cdot \cos b}{\cos a \cdot \cos b / \cos a \cdot \cos b + \sin a \cdot \sin b / \cos a \cdot \cos b} = \frac{\sin a / \cos a + \sin b / \cos b}{1 + \tan a \cdot \tan b} = \\ = \frac{\tan a + \tan b}{1 + \tan a \cdot \tan b}. \end{array}$$

Proved.

Answer: proved.