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

(sec A - cosec A)(1 + tan A + cot A) = tan A sec A - cot A cosec A.

Solving L.H.S.,

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

Now solving R.H.S.,

$\Rightarrow \text{tan A sec A - cot A cosec A} \\[1em] = \dfrac{\text{sin A}}{\text{cos A}} \times \dfrac{1}{\text{cos A}} - \dfrac{\text{cos A}}{\text{sin A}} \times \dfrac{1}{\text{sin A}} \\[1em] = \dfrac{\text{sin A}}{\text{cos}^2 A} - \dfrac{\text{cos A}}{\text{sin}^2 A} \\[1em] = \dfrac{\text{sin}^3 A - \text{cos}^3 A}{\text{sin}^2 A \text{ cos}^2 A} \\[1em] = \dfrac{\text{(sin A - cos A)(sin}^2 A + \text{cos}^2 A + \text{sin A cos A})}{\text{sin}^2 A \text{ cos}^2 A} \\[1em] = \dfrac{(\text{sin A - cos A})(\text{sin A cos A + 1})}{\text{sin}^2 A \space \text{cos}^2 A}.$

Since, L.H.S. = R.H.S. hence proved that (sec A - cosec A)(1 + tan A + cot A) = tan A sec A - cot A cosec A.