# How do you prove ( tan A + tan B) / (1-tan A tan B) = (cot A + cot B ) / ( cot A cot B - 1)?

Apr 9, 2018

We can rewrite the right hand side as follows:

$= \frac{\frac{1}{\tan} A + \frac{1}{\tan} B}{\frac{1}{\tan} A \frac{1}{\tan} B - 1}$

=((tanB + tanA)/(tanAtanB))/((1 - tanAtanB)/(tanAtanB)

$= \frac{\tan A + \tan B}{1 - \tan A \tan B}$

Since the left and right hand sides are equal, the identity has been proved.

Hopefully this helps!