Prove the identity: $\dfrac{\tan \left(x - \dfrac{\pi}{2}\right) \cos \left(\dfrac{3 \pi}{2} + x\right) - \sin^3 \left(\dfrac{7 \pi}{2} - x\right)}{\cos \left(x - \dfrac{\pi}{2}\right) \tan \left(\dfrac{3 \pi}{2} + x\right)} = \sin^2 x$
$\begin{aligned}
\tan \left(x - \dfrac{\pi}{2}\right) & = \tan \left[- \left(\dfrac{\pi}{2} - x\right)\right] \\\\
& = - \tan \left(\dfrac{\pi}{2} - x\right) \\\\
& = - \cot x \\\\
& = \dfrac{- \cos x}{\sin x} \;\;\; \cdots \; (1)
\end{aligned}$
$\begin{aligned}
\cos \left(\dfrac{3 \pi}{2} + x\right) & = \cos \left[\pi + \left(\dfrac{\pi}{2} + x\right)\right] \\\\
& = -\cos \left(\dfrac{\pi}{2} + x\right) \\\\
& = - \left(- \sin x\right) \\\\
& = \sin x \;\;\; \cdots \; (2)
\end{aligned}$
$\begin{aligned}
\sin \left(\dfrac{7 \pi}{2} - x\right) & = \sin \left[\dfrac{\pi}{2} + \left(3 \pi - x\right)\right] \\\\
& = \cos \left(3 \pi - x\right) \\\\
& = - \cos x \;\;\; \cdots \; (3)
\end{aligned}$
$\begin{aligned}
\cos \left(x - \dfrac{\pi}{2}\right) & = \cos \left[- \left(\dfrac{\pi}{2} - x\right)\right] \\\\
& = \cos \left(\dfrac{\pi}{2} - x\right) \\\\
& = \sin x \;\;\; \cdots \; (4)
\end{aligned}$
$\begin{aligned}
\tan \left(\dfrac{3 \pi}{2} + x\right) & = \tan \left[\pi + \left(\dfrac{\pi}{2} + x\right)\right] \\\\
& = \tan \left(\dfrac{\pi}{2} + x\right) \\\\
& = \dfrac{- \cos x}{\sin x} \;\;\; \cdots \; (5)
\end{aligned}$
$\begin{aligned}
\therefore \; LHS & = \dfrac{\tan \left(x - \dfrac{\pi}{2}\right) \cos \left(\dfrac{3 \pi}{2} + x\right) - \sin^3 \left(\dfrac{7 \pi}{2} - x\right)}{\cos \left(x - \dfrac{\pi}{2}\right) \tan \left(\dfrac{3 \pi}{2} + x\right)} \\\\
& = \dfrac{\dfrac{- \cos x}{\sin x} \times \sin x - \left(- \cos x\right)^3}{\sin x \times \left(\dfrac{- \cos x}{\sin x}\right)} \;\;\; \left[\text{In view of equations } (1) \cdots (5) \right] \\\\
& = \dfrac{- \cos x + \cos^3 x}{- \cos x} \\\\
& = 1 - \cos^2 x \\\\
& = \sin^2 x = RHS
\end{aligned}$
Hence proved.