I'm trying to evaluate this limit and i'm having some issues.

$\displaystyle \lim_{x\to 0} \frac{1-\sec^2(2x)}{x^2} $

I'm not 100% sure about my trig Identities but i think that

$\displaystyle 1-\sec^2(2x) $ is equal to $\displaystyle 2\tan^2(x) $

is that correct?

if so, i say that the equation is equal to $\displaystyle \frac{2\sin^2(x)}{2\cos^2(x)}(\frac{1}{x^2}) $

I have a feeling this is wrong somehow knowing the answer is -4. Can someone help me?

thank you