Prove that (sec2 theta)=(sec^2 theta)/(2-sec^2 theta)
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Prove that (sec2 theta)=(sec^2 theta)/(2-sec^2 theta)

[From: ] [author: ] [Date: 11-11-18] [Hit: ]
just to clarify, you multiplied the top and bottom by sec^2 theta for that third step?......
Really stuck on proving this trig identity. I know that I should probably use the double angle formula, but I end up getting really complicated answers. Any help would be appreciated! Thanks =)

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Show: sec(2Ɵ) = sec²(Ɵ) / (2 - sec²Ɵ)

sec(2Ɵ) = 1 / cos(2Ɵ) = 1 / (2cos²Ɵ - 1) = sec²Ɵ / ((2cos²Ɵ - 1) (sec²Ɵ)) = sec²Ɵ / (2 - sec²Ɵ)

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wait, just to clarify, you multiplied the top and bottom by sec^2 theta for that third step?

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