Question

5A=tan(3A+2A)=tan3A+Tan2A/1-Tan3A*Tan2A prove it.

Answer 1

Tan(3A+2A)= tan3A+tan2A /1-tan3A.tan2A =0,

Answer 2

I think that it should be Tan5A=Tan(3A+2A) =Tan3A+Tan2A/1-Tan3A*Tan2A

Then,

Tan5A=tan3A+Tan2A/1-Tan3A*Tan2A

=Tan5A(1-Tan3A*Tan2A) =Tan3A+Tan2A

Or=Tan5A-Tan5A*Tan3A*Tan2A=Tan3A, +Tan2A.