if 2+2tanA=sec^2a, find the value of TanA.?

sec^2 A = 1 + tan^2 A

2 + 2tan A - 1 - tan^2 A = 0

tan^2 A - 2tan A - 1 = 0

tan^2 A - 2tan A + 1 - 1 - 1 = 0

(tan A - 1 )^2 = 2

tan A - 1 = 1.414
tan A = 2.414
A = 67.498 degrees, 247.498 degrees

--------------OR
tan A - 1 = -1.414
tan A = 0.414
A = 22.490 degrees, 202.490 degrees

Edit
I used method of completing the squares to solve the equation. Let me explain the simpler one, the quadratic formula. Say
tan A = (- b+/-Sqrt[b^2 - 4ac])/2a, where a = 1, b = -2 and c = 1. a and b are coefficients of tan^2 A and tan A respectively while c is the constant.

Method of completing the squares which I used will result in the same answer. But there is no point explaining that to you, I suggest you understand the basic quadratic formula first. I hope I have made it clear.

Good luck

2 + 2 tan A = sec^2 A
2 + 2 tan A = 1 + tan^2 A
0 = tan^2 A - 2 tan A - 1
tan A =1 ± √2

2 + 2tanA = sec^2A
2 + 2tanA = 1 + tan^2A (divide sin^2A + cos^2A = 1 by cos^2A)
1 + 2tanA = tan^2A
tan^2A - 2tanA - 1 = 0
tanA = [2 ± √(4 + 4)]/2 = 1 ± √2 (quadratic formula)

Quadratic formula:
For ax^2 + bx + c = 0, the value of x is given by:
x = [-b ± √(b^2 - 4ac)]/(2a)