Help doing a few trig proofs

We will use basic trigonometric identities along the way. 
i) 
\[\frac{(1-\tan^2x)}{\sin2x }= \frac{(1-\frac{\sin^2 x}{\cos ^2 x})}{\sin2x }=\frac{\frac{\cos^2 x -\sin^2 x}{\cos ^2 x}}{\sin2x }\]
\[=\frac{\frac{\cos 2x}{\cos ^2 x}}{\sin2x }=\frac{\frac{1}{\cos ^2 x}}{\frac{\sin2x}{\cos 2x} }=\frac{\sec^2 x}{\tan 2x}.\]
ii)
\[\frac{\sin 2x}{\cot^2 x}=\frac{2 \sin x \cos x}{\frac{\cos ^2 x}{\sin ^2 x}}=\frac{2 \sin^3 x}{\cos x}=\frac{\sin x}{\cos x}(2 \sin^2 x) \]
\[=\tan x (1-\cos 2x).\]

(iii) 
\[\frac{\sec^2 x}{1-\tan^2 x}=\frac{\frac{1}{\cos^2 x}}{1-\frac{\sin^2 x}{\cos^2 x}}\]
\[=\frac{\frac{1}{\cos^2 x}}{\frac{\cos^2 x -\sin^2 x}{\cos^2 x}}=\frac{1}{\cos^2 x -\sin^2 x}\]
\[=\frac{1}{\cos 2x}=\frac{\sin 2x}{\cos 2x}\frac{1}{\sin 2x}=\tan 2x\frac{1}{\sin 2x} \]
\[=\frac{\tan 2x}{\sin 2x}.\]