Prove that :

(cosec A - sin A)(sec A - cos A) sec² A = tan A

Solving L.H.S. of the equation :

$\Rightarrow \Big(\dfrac{1}{\text{sin A}} - \text{sin A}\Big) \Big(\dfrac{1}{\text{cos A}} - \text{cos A}\Big)\text{sec}^2 \text{A} \\[1em] \Rightarrow \Big(\dfrac{\text{1 - sin}^2 A}{\text{sin A}}\Big) \times \Big(\dfrac{\text{1 - cos}^2 A}{\text{cos A}}\Big) \times \text{sec}^2 A$

By formula,

1 - sin² A = cos² A and 1 - cos² A = sin² A

$\Rightarrow \dfrac{\text{cos}^2 A}{\text{sin A}} \times \dfrac{\text{sin}^2 A}{\text{cos A}} \times \dfrac{1}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{tan A}.$

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

Hence, proved that (cosec A - sin A)(sec A - cos A) sec² A = tan A