Trigonometry

Prove the identity: $\;$ $\dfrac{\tan^2 \theta}{\tan^2 \theta - 1} + \dfrac{cos^2 \theta}{\sin^2 \theta - \cos^2 \theta} = \dfrac{1}{1 - 2 \cos^2 \theta}$


$\begin{aligned} LHS & = \dfrac{\tan^2 \theta}{\tan^2 \theta - 1} + \dfrac{cos^2 \theta}{\sin^2 \theta - \cos^2 \theta} \\\\ & = \dfrac{\dfrac{\sin^2 \theta}{\cos^2 \theta}}{\dfrac{\sin^2 \theta}{\cos^2 \theta} - 1} + \dfrac{cos^2 \theta}{\sin^2 \theta - \cos^2 \theta} \\\\ & = \dfrac{\sin^2 \theta}{\sin^2 \theta - \cos^2 \theta} + \dfrac{cos^2 \theta}{\sin^2 \theta - \cos^2 \theta} \\\\ & = \dfrac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta - \cos^2 \theta} \\\\ & = \dfrac{1}{\sin^2 \theta - \cos^2 \theta} \\\\ & = \dfrac{1}{1 - \cos^2 \theta - \cos^2 \theta} \\\\ & = \dfrac{1}{1 - 2 \cos^2 \theta} = RHS \end{aligned}$

Hence proved.