Question 749317
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.