Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

(cosec A – sin A)(sec A – cos A) = $\dfrac{1}{\text{tan A + cot A}}$

To prove:

(cosec A – sin A)(sec A – cos A) = $\dfrac{1}{\text{tan A + cot A}}$

Solving L.H.S. of the given equation,

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

Solving R.H.S. of the given equation,

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

Since, L.H.S. = R.H.S. = sin A cos A

Hence, proved that (cosec A – sin A)(sec A – cos A) = $\dfrac{1}{\text{tan A + cot A}}$.