# if cos2A=tan²x, then prove cos2x=tan²A

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You need to prove that `cos 2x = tan^2 A` , using `cos 2A = tan^2 x` such that:

You need to add 1 both sides such that:

`cos 2x + 1= tan^2 A + 1`

You need to remember that `cos 2x = 2cos^2 x - 1` such that:

`2cos^2 x - 1 + 1 = tan^2 A + 1`

`2cos^2 x = tan^2 A + 1`

You should substitute `1/(cos^2 A)` for `tan^2 A + 1` such that:

`2cos^2 x = 1/(cos^2 A) =gt 2cos^2 A= 1/(cos^2 x)`

You need to substitute `tan^2x + 1` for `1/(cos^2 x)` such that:

`2cos^2 A = tan^2 x + 1`

You need to subtract 1 both sides such that:

`2cos^2 A - 1 = tan^2 x + 1 - 1`

You need to substitute `cos 2A` for `2cos^2 A - 1` such that:

`cos 2A = tan^2 x`

**Since the problem provides the information that `cos 2A = tan^2 x` , then the last line, that is the equivalent to `cos 2x = tan^2 A` , proves that`cos 2x = tan^2 A` is true.**

**Sources:**

I got the answer...

cos2a = tan²x

(1-tan²A)/(1+tan²A) = tan²x

1-tan²A = tan²x + tan²A.tan²x

1-tan²x = tan²A.tan²x + tan²A

1-tan²x = tan²A(tan²x + 1)

(1-tan²x)/(1+tan²x) = tan²A

**cos2x = tan²A**

Given: cos2A=tan²x

=> cos²A - sin²A = tan²x

=> cos²A - (1 - cos²A) = tan²x

=> 2cos²A - 1 = tan²x

=> 2cos²A = tan²x + 1

=> 2cos²A = sec²x

=> 2cos²A = 1/cos²x

=> 2cos²x = 1/cos²A

=> 2cos²x = sec²A

=> 2cos²x - 1 = sec²A - 1

=> cos²x - (1 - cos²x) = tan²A

=> cos²x - sin²x = tan²A

=> cos2x = tan²A (Hence, Proved)