Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

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

L.H.S. of the equation can be written as,

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

As 1 - sin² A = cos² A and 1 - cos² A = sin² A.

$\Rightarrow \dfrac{\text{cos}^2 A \text{ sin}^2 A}{\text{sin A cos A}} \times \dfrac{1}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A \text{ sin}^2 A}{\text{cos}^3 A \text{ sin} A} \\[1em]$

Dividing numerator and denominator by sin A cos A.

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

Since, L.H.S. = R.H.S. hence proved that, (cosec A - sin A)(sec A - cos A)sec² A = tan A.