Derive tan(a+b), using sin (a +b) and cos(a +b).

To derive t(a+b) , using  sin(a+b) and
cos(a+b).

We know the identities (i) sin(a+b) =
sina*cosn+cosa*sinb and

(ii) cos(a+b) =
csa*cos-sina*sinb.

Therefore tan(a+b) =
sin(a+b)/cos(a+b).....(1).

We substitute sin(a+b) =
sina*cosb+cos*sinb and cos(a+b) = cosa*cosb-sina*sin b in
(1):

 tan (a+b) =
(sina*cosb+cosa*sinb)/(cosa*cos-sina*sinb. We divide  both numerator and denominator in
right by cosa*cosb and get:

tan(a+b) =
{cosa*sinb/(cosa*cosb) - sina*cosb/(coa*cosb)}/{cosa*cosb/(cosacosb) -
sina*sinb/(cosa*cosb)}

Therefore tan(a+b) =
(tana + tanb)/(1- tana*tan*b).