Hello candidate,

First of all, we can replace B as π/2- 2A, or A as π/4- B/2. So, now upon commutating the value of tan(A+B), we get the result as tanA + tan (π/2-2A)/ 1- tanA* tan(90- 2A).

Upon solving the equation, 2 TanA+ tanB, we get the same value as tanA + tan (π/2-2A)/ 1- tanA* tan(90- 2A). As the right hand side part is same in both the cases, hence both the equations are same.

Hope it helps!!