Prove the identity: $\dfrac{\cot \alpha + \cot \left(270^\circ + \alpha\right)}{\cot \alpha - \cot \left(270^\circ + \alpha\right)} + 2 \cos \left(135^\circ + \alpha\right) \cos \left(135^\circ - \alpha\right) = 2 \cos 2 \alpha$
$\begin{aligned}
LHS & = \dfrac{\cot \alpha + \cot \left(270^\circ + \alpha\right)}{\cot \alpha - \cot \left(270^\circ + \alpha\right)} + 2 \cos \left(135^\circ + \alpha\right) \cos \left(135^\circ - \alpha\right) \\\\
& = \dfrac{\cot \alpha - \tan \alpha}{\cot \alpha + \tan \alpha} + 2 \cos \left(\dfrac{270^\circ + 2 \alpha}{2}\right) \cos \left(\dfrac{270^\circ - 2 \alpha}{2}\right) \\\\
& = \dfrac{\cot \alpha - \tan \alpha}{\cot \alpha + \tan \alpha} + \left[\cos \left(270^\circ\right) + \cos 2 \alpha\right] \\\\
& = \dfrac{\dfrac{1}{\tan \alpha} - \tan \alpha}{\dfrac{1}{\tan \alpha} + \tan \alpha} + 0 + \cos 2 \alpha \\\\
& = \dfrac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} + \cos 2 \alpha \\\\
& = \dfrac{1 - \dfrac{\sin^2 \alpha}{\cos^2 \alpha}}{1 + \dfrac{\sin^2 \alpha}{\cos^2 \alpha}} + \cos 2 \alpha \\\\
& = \dfrac{\cos^2 \alpha - \sin^2 \alpha}{\cos^2 \alpha + \sin^2 \alpha} + \cos 2 \alpha \\\\
& = \dfrac{\cos 2 \alpha}{1} + \cos 2 \alpha \\\\
& = 2 \cos 2 \alpha = RHS
\end{aligned}$
Hence proved.