Prove the following identity:

tan² A - sin² A = sin² A tan² A

Solving L.H.S of equation,

$\Rightarrow \dfrac{\sin^2 A}{\cos^2 A} - \sin^2 A \\[1em] \Rightarrow \sin^2 A \Big( \dfrac{1}{\cos^2 A} - 1 \Big) \\[1em] \Rightarrow \sin^2 A \Big( \dfrac{1 - \cos^2 A}{\cos^2 A} \Big) \\[1em] \Rightarrow \sin^2 A \Big(\dfrac{\sin^2 A}{\cos^2 A} \Big) \\[1em] \Rightarrow \sin^2 A \tan^2 A.$

Since, L.H.S. = R.H.S.,

Hence, proved that tan² A - sin² A = sin² A tan² A.