A triangle ABC is right angled at B; find the value of $\dfrac{\text{sec A. cosec C - tan A. cot C}}{\text{sin B}}$

In triangle ABC,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + ∠C = 180°

⇒ ∠A + ∠C = 90°

⇒ ∠A = 90° - ∠C.

Substituting value of A in $\dfrac{\text{sec A. cosec C - tan A. cot C}}{\text{sin B}}$ we get,

$\Rightarrow \dfrac{\text{sec (90° - C). cosec C - tan (90° - C). cot C}}{\text{sin 90°}}$

By formula,

tan (90° - C) = cot c, sec (90° - C) = cosec C and cosec² C - cot² C = 1.

$\Rightarrow \dfrac{\text{cosec C. cosec C - cot C. cot C}}{1} \\[1em] \Rightarrow \text{cosec}^2 C - \text{cot}^2 C \\[1em] \Rightarrow 1.$

Hence, $\dfrac{\text{sec A. cosec C - tan A. cot C}}{\text{sin B}}$ = 1.