Question

tan(45+A)=1+tanA/1-tanA​

Answer 1

Answer:

by using the identity tan(A+B)

Step-by-step explanation:

by taking LHS

tan(45+A)

Answer 2

Use tangent addition to prove that tan(A + 45) = (1 + tan(A))/(1 - tan(A))...

tan(α + ß) = (tanα + tanß)/(1 - tanαtanß)

So...

LHS
= tan(A + 45) = (tan(A) + tan(45))/(1 - tan(A)tan(45))
= (tan(A) + 1)/(1 - tan(A)(1))
= (1 + tan(A))/(1 - tan(A))
= RHS

I hope this helps!