Take left hand side of the equation (1+tanA)2+(1−tanA)2=2sec2A.
LHS = (1+tanA)2+(1−tanA)2=1+tan2A+2tanA+1+tan2A−2tanA
=2+2tan2A
=2(1+tan2A)
=2sec2A
=RHS
(1+tan A)2+(1−tan A)2=