Question

Can anyone prove to me:

Answer 1

\[\tan(a+b) = \frac{\tan(a)+\tan(b)}{1 - \tan(a)\tan(b)}\]

Answer 2

im actually on this assignment also..

Answer 3

It's not an assignment. I'm just curious because I can't find it with sin(a+b) and cos(a+b) identity

Answer 4

how about writing it in the form\[\tan(a+b)=\frac{\sin(a+b)}{\cos(a+b)}\]

Answer 5

tried

Answer 6

I can't get to that tan identity

Answer 7

emm...\[\tan(a+b)=\frac{\sin(a+b)}{\cos(a+b)}=\frac{\sin a \cos b+\cos a \sin b}{\cos a \cos b-\sin a \sin b}\]divide num and denum by \(\cos a \cos b\)