Prove that (1-tan squared x)/(1+tan squared x) = cos2x?

Answer 1

[1 - tan^2(x)] / [1 + tan^2(x)]

tan(x) = sin(x)/cos(x)

[1 - sin^2(x)/cos^2(x)] / [1 + sin^2(x)/cos^2(x)]

[cos^2(x) - sin^2(x)]/cos^2(x) / {[cos^2(x) + sin^2(x)]/cos^2(x)}

[cos^2(x) - sin^2(x)]/[cos^2(x) + sin^2(x)]

Use the identity cos^2(x) + sin^2(x) = 1

cos^2(x) - sin^2(x)

And cos(2x) = cos^2(x) - sin^2(x) is an identity that completes the proof.

Answer 2

=> (1-tan ^2 x)/sec^2 x
=> {1-(sin^2 x /cos^2 x)} / {1/cos^2 x}
=> cos^2 x - sin^2 x
=> cos2x

hope u r satisfied... sin^2 x=sin square x (the same is for tan& cos)

Answer 3

N = 1 - tan ² x
N = 1 - sin ² x / cos ² x
N = (cos ² x - sin ² x) / cos ² x
N = cos 2x / cos ² x

D = 1 + tan ² x
D = 1 + sin ² x / cos ² x
D = (cos ² x + sin ² x) / cos ² x
D = 1 / cos ² x

N/D = cos 2x

Answer 4

since 1 + tan^2(x) = sec^2(x)

[1- tan^2(x)]/(1+tan^2(x)) = [(1 - tan^2(x)]/sec^2(x)

=> [1 / sec^2(x)] - [tan^2(x)/sec^2(x)]

=> cos^2(x) - sin^2(x)

=> cos(2x)

Answer 5

wow