Prove that :

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

Solving L.H.S. of the equation :

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

By formula,

1 - sin² A = cos² A

1 - cos² A = sin² A

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

Solving R.H.S. of the equation :

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

By formula,

sin² θ + cos² θ = 1

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

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

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