Trigonometry - Identity Transformation of Trigonometric Expressions

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


$\begin{aligned} LHS & = \dfrac{1 - 2 \sin^2 \alpha}{1 + \sin 2 \alpha} \\\\ & = \dfrac{\cos 2 \alpha}{1 + \sin 2 \alpha} \\\\ & = \dfrac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} \div \left(1 + \dfrac{2 \tan \alpha}{1 + \tan^2 \alpha}\right) \\\\ & = \left(\dfrac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha}\right) \times \left(\dfrac{1 + \tan^2 \alpha}{1 + \tan^2 \alpha + 2 \tan \alpha}\right) \\\\ & = \dfrac{\left(1 + \tan \alpha\right) \left(1 - \tan \alpha\right)}{\left(1 + \tan \alpha\right)^2} \\\\ & = \dfrac{1 - \tan \alpha}{1 + \tan \alpha} = RHS \end{aligned}$

Hence proved.