In △ ABC, ∠B = 90°, evaluate : cosec A cos C - sin A sec C.

In triangle ABC, sum of all angles is 180°.

∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + ∠C = 180°

⇒ ∠A + ∠C = 180° - 90°

⇒ ∠A + ∠C = 90°

⇒ ∠C = 90° - ∠A

Given: cosec A cos C - sin A sec C

$= \dfrac{1}{\text{sin A}}\text{cos C} - \text{sin A}\dfrac{1}{\text{cos C}}\\[1em] = \dfrac{\text{cos C}}{\text{sin A}} - \dfrac{\text{sin A}}{\text{cos C}}\\[1em] = \dfrac{\text{cos (90° - A)}}{\text{sin A}} - \dfrac{\text{sin A}}{\text{cos (90° - A)}}\\[1em] = \dfrac{\text{sin A}}{\text{sin A}} - \dfrac{\text{sin A}}{\text{sin A}}\\[1em] = 1 - 1\\[1em] = 0$

Hence, the value of cosec A cos C - sin A sec C = 0.