How do I verify this Trigonometric Identity?

Question

I can't figure out whether to start on the right or left hand side. Anyways here is the question:

$\sin\theta + \cos\theta = \frac{1-2\cos^2\theta}{\sin\theta-\cos\theta}$

Answer 1

\begin{align} \dfrac{1-2\cos^2 \theta}{\sin \theta-\cos \theta} &= \dfrac{1-\cos^2\theta-\cos^2\theta}{\sin \theta-\cos \theta} \\ &= \dfrac{\sin^2\theta-\cos^2\theta}{\sin \theta-\cos \theta} \\ &= \dfrac{(\sin \theta-\cos \theta)(\sin \theta+\cos \theta)}{\sin \theta-\cos \theta} \\ &= \sin \theta+\cos \theta \end{align}

Answer 2

Use cross-products. $$\sin\theta+\cos\theta = \frac{1-2\cos^2\theta}{\sin\theta-\cos\theta} \implies (\sin\theta+\cos\theta)(\sin\theta-\cos\theta) = 1-2\cos\theta$$ $$\sin^2\theta-\cos^2\theta = 1-2\cos^2\theta$$ $$1-\cos^2\theta-\cos^2\theta = 1-2\cos^2\theta \implies \boxed{1-2\cos^2\theta = 1-2\cos^2\theta}$$ Verified.