how is tan (A+B) = [tan A + tan B]/[1 - tan A tan B]?

How is tan (A+B) = [tan A + tan B]/

[1 - tan A tan B]?

Method 1:

Let A = 30 deg and B = 45 deg.

LHS = tan (30+45) = tan 75 = 3.732050808

RHS = [tan A + tan B]/[1 - tan A tan B]

= [tan 30 + tan 45]/[1 - tan 30 tan 45]

= [0.577350269 + 1]/[1 - 0.577350269*1]

= 1.577350269/0.42264973

= 3.732050808 = LHS

Proved.

Method 2:

tan (A+B) = [tan A + tan B]/[1 - tan A tan B]

RHS = [tan A + tan B]/[1 - tan A tan B]

=[(sin A/cos A) + (sin B/cos B)]/[1-(sin A/cos A)(sin B/cos B)

= [sin A cos B + cos A sin B]/[cos A cos B][1 - sin A sin B/(cos A cos B)]

= sin (A+B)/{[cos A cos B][cos A cos B - sin A sin B]/(cos A cos B)}

= sin (A+B)/[cos A cos B - sin A sin B

= sin (A+B)/cos (A+B)

= tan (A+B) = LHS.

Proved.

Thanks.

source:Quara