SOLUTION: 1/(sec(θ)-tan(θ))=sec(θ)+tan(θ) how do I prove this?

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Question 749317: 1/(sec(θ)-tan(θ))=sec(θ)+tan(θ) how do I prove this?
Answer by FrankM(1040) About Me  (Show Source):
You can put this solution on YOUR website!
1/(sec(θ)-tan(θ))=sec(θ)+tan(θ)

(sec(θ)-tan(θ)) (sec(θ)+tan(θ)) = 1 (we cross multiplied)

(sec^2(θ)-tan^2(θ)) = 1 ( (a+b) * (a-b) = a^2 - b^2 )

tan^2(θ) + 1 = sec^2(θ) (rearranged)

sin^2(θ)/cos^2(θ) + cos^2(θ)/cos^2(θ)= 1/cos^2(θ)

[Here, Tan = Sin/Cos, Substitute Cos/Cos for 1 and 1/Cos for Sec]

sin^2(θ)+cos^2(θ)^2=1 (multiply both sides by cos^2)

Above equation is the pythagorean theorum. Solved.