Prove the identity: $\dfrac{\cos^2 \beta \left(\tan^2 \alpha - \tan^2 \beta\right)}{\sec^2 \alpha} = \sin \left(\alpha + \beta\right) \sin \left(\alpha - \beta\right)$
$\begin{aligned}
LHS & = \dfrac{\cos^2 \beta \left(\tan^2 \alpha - \tan^2 \beta\right)}{\sec^2 \alpha} \\\\
& = \cos^2 \alpha \cos^2 \beta \left(\dfrac{\sin^2 \alpha}{\cos^2 \alpha} - \dfrac{\sin^2 \beta}{\cos^2 \beta}\right) \\\\
& = \dfrac{\cos^2 \alpha \cos^2 \beta \left(\sin^2 \alpha \cos^2 \beta - \sin^2 \beta \cos^2 \alpha\right)}{\cos^2 \alpha \cos^2 \beta} \\\\
& = \sin^2 \alpha \cos^2 \beta - \sin^2 \beta \cos^2 \alpha \\\\
& = \left(\sin \alpha \cos \beta\right)^2 - \left(\sin \beta \cos \alpha\right)^2 \\\\
& = \left(\sin \alpha \cos \beta + \sin \beta \cos \alpha\right) \left(\sin \alpha \cos \beta - \sin \beta \cos \alpha\right) \\\\
& = \sin \left(\alpha + \beta\right) \sin \left(\alpha - \beta\right) = RHS
\end{aligned}$
Hence proved.