How do I verify that this is an identitiy?

Question

1 + cos2A = 2 / (1 + tan^2 A)

Answer 1

1+cos2A=2/(1/cos^2A).....'cuz 1+tan^2A=1/cos^2A
1+cos^2A-sin^2A = 2cos^2A ......... 'cuz cos2A=cos^2A-sin^2A
cos^2A + cos^2A = 2cos^2A ......... 'cuz 1-sin^2A = cos^2A
2cos^2A = 2cos^2A

Answer 2

Ok, let's try this one...

tan(A) = sin(A)/Cos(A)

1 + tan^2(A) = 1+ sin^2(A)/cos^2(A)
= (cos^2(A) + sin^2(A))/cos^2(A)
= 1/cos^2(A)

2/(1/cos^2(A)) = 2cos^2(A)

now, working with the 1 + cos2A,

cos(A+A) = cos(A)cos(A) - sin(A)sin(A) = cos^2(A) - sin^2(A)
and
1 = cos^2(A) + sin^2(A)

1+ cos2A = 2 cos^2(A)

Good luck!

Answer 3

1 + cos(2A) = 2/(1 + tan(A)^2)
1 + cos(2A) = 2/(1 + (sinA/cosA)^2)
1 + cos(2A) = 2/((cos(A)^2 + sin(A)^2)/(cos(A)^2))
1 + cos(2A) = 2/(1/cos(A)^2)
1 + cos(2A) = 2cos(A)^2

cos(2A) = 2cos(A)^2 - 1

1 + cos(2A) = 2cos(A)^2
cos(2A) = 2cos(A)^2 - 1

Answer 4

1+2cos^A-1
=2cos^2A
=2/sec^2A
=2/(1+tan^2A)