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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
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