I'm wondering how to solve the following limit: $$\lim\limits_{x\to1}\frac{\tan(x^2-1)}{\sin(x^2-4x+3)}$$ using only basic trigonometric identities from precalculus and the basic trigonometric limits $\lim\limits_{\theta\to0}\frac{\sin\theta}{\theta} = 1$ and $\lim\limits_{\theta\to0}\frac{1-\cos\theta}{\theta} = 0$, i.e. without using L'Hôpital's rule or Taylor expansions.
I've made some headway using the substitution $u = x-1$ after some general simplifying, but cannot get a clean result.