(tan A + cot A)(cosec A - sin A)(sec A - cos A) = 1

Solving L.H.S. of the above equation :

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

By formula,

sin² A + cos² A = 1, 1 - sin² A = cos² A and 1 - cos² A = sin² A.

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

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

Hence, proved that (tan A + cot A)(cosec A - sin A)(sec A - cos A) = 1.