Prove the identity: $\dfrac{1}{4 \sin^2 \alpha \cos^2 \alpha } - \dfrac{\left(1 - \tan^2 \alpha\right)^2}{4 \tan^2 \alpha} = 1$
To Prove That (TPT) $\;$ $\dfrac{1}{4 \sin^2 \alpha \cos^2 \alpha } - \dfrac{\left(1 - \tan^2 \alpha\right)^2}{4 \tan^2 \alpha} = 1$
i.e. $\;$ TPT $\;$ $1 + \dfrac{\left(1 - \tan^2 \alpha\right)^2}{4 \tan^2 \alpha} = \dfrac{1}{4 \sin^2 \alpha \cos^2 \alpha}$
$\begin{aligned}
LHS & = 1 + \dfrac{\left(1 - \tan^2 \alpha\right)^2}{4 \tan^2 \alpha} \\\\
& = \dfrac{4 \tan^2 \alpha + 1 + \tan^4 \alpha - 2 \tan^2 \alpha}{4 \tan^2 \alpha} \\\\
& = \dfrac{1 + \tan^4 \alpha + 2 \tan^2 \alpha}{4 \tan^2 \alpha} \\\\
& = \dfrac{\left(1 + \tan^2 \alpha\right)^2}{4 \tan^2 \alpha} \\\\
& = \dfrac{\left(\sec^2 \alpha\right)^2}{4 \tan^2 \alpha} \\\\
& = \dfrac{1}{\cos^4 \alpha} \times \dfrac{\cos^2 \alpha}{4 \sin^2 \alpha} \\\\
& = \dfrac{1}{4 \sin^2 \alpha \cos^2 \alpha} = RHS
\end{aligned}$
Hence proved.