Prove the identity sec 2Θ = (sec^2 Θ) / (2 - sec^2 Θ)?

Question

Having trouble with this problem on my study guide, and would appreciate the steps on how to derive the answer to prepare me for my upcoming test. Thanks in advance for anyone reading this and trying to help out.

Answer 1

I'm going to leave the left side alone, and change just the right side, until the right side looks like the left side. Before I start, here are some important trig. identities that I will use:

sec² Θ = 1 + tan² Θ
cos 2Θ = cos² Θ - sin² Θ


Now, I will begin the proof:

sec 2Θ = (sec² Θ) / (2 - sec² Θ)

= (1/cos² Θ) / (2 - sec² Θ)

= (1/cos² Θ) / [2 - (1 + tan² Θ)]

= (1/cos² Θ) / (1 - tan² Θ)

= 1 / [(cos² Θ)( 1 - tan² Θ)]

= 1 / [cos² Θ - sin² Θ]

= 1 / cos 2Θ

= sec 2Θ

Therefore...:

sec 2Θ = sec 2Θ (this MUST be true, right?)

Q.E.D.

Answer 2

RHS = (1 / cos² x) / (2 - 1 / cos²x)
RHS = (1 / cos²x) / (2cos²x - 1) / cos²x
RHS = 1 / (2 cos²x - 1)
RHS = 1 / cos 2x
RHS = sec 2x = LHS
Sorry I can`t find theta!

Answer 3

sec (2x) = (sec^2 (x)) / (2 - sec^2 (x))?
Take the reciprocal
cos(2x) =(2 -sec^2(x) )/(sec^2(x) )
Use the identity cos(2x) = 2cos^2(x) -1
2cos^2(x) -1 =(2 -sec^2(x) )/(sec^2(x) )
Multiply both sides by sec^2(x)
sec^2(x)*(2cos^2(x) -1) =2 -sec^2(x)
2 -sec^2(x) =2 -sec^2(x)