Prove that tan(a+b)=tana+tanb/1-tana*tanb

First, you need to write this expression correctly, using
brackets:

tan(a+b) = (tan a + tan b)/(1 - tan a*tan
b)

We'll prove this identity, using the information that
tangen function is a ratio:

tan (a+b) = sin (a+b)/cos
(a+b)

We'll use the following
identities:

sin (a+b) = sin a*cos b + sin b*cos
a

cos (a+b) = cos a*cos b - sin a*sin
b

tan (a+b) = (sin a*cos b + sin b*cos a)/(cos a*cos b -
sin a*sin b)

We'll force factor cos a*cos b, both numerator
and denominator, creating the tangent functions within
brackets:

tan (a+b) = cos a*cos b( tan a + tan b)/cos a*cos
b(1 - tan a*tan b)

We'll simplify and we'll
get:

tan (a+b) = (tan a + tan b)/(1 - tan a*tan
b)

Therefore, the identity tan (a+b) = (tan a
+ tan b)/(1 - tan a*tan b) is verified.