Question

(1+tanA)^2+(1-tanA)^2=2sec^2A

Answer 1

TAKING LHS

(1+tanA)^2+(1-tanA)^2

1+tan^2A+2tanA+1+tan^2A-2tanA           [using(a+b)^2and (a-b)^2]

2+2tan^2A

2(1+tan^2A)      

2Sec^2A          (1+tan^2A=sec^2A)

SO LHS=RHS

HENCE PROVED