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

Given:

ABC is right-angled triangle at B.

∠ A + ∠ B + ∠ C = 180°

⇒ ∠ A + 90° + ∠ C = 180°

⇒ ∠ A + ∠ C = 180° - 90°

⇒ ∠ A + ∠ C = 90°

⇒ ∠ A = 90° - ∠ C

Now,

$\dfrac{\text{sec A . sin C - tan A . tan C}}{\text{sin B}}\\[1em] = \dfrac{\text{sec (90° - C) . sin C - tan (90° - C). tan C}}{\text{sin B}}\\[1em] = \dfrac{\text{cosec C . sin C - cot C. tan C}}{\text{sin B}}\\[1em] = \dfrac{\dfrac{1}{\text{sin C}} \times \text{sin C} - \dfrac{1}{\text{tan C}} \times \text{tan C}}{\text{sin B}}\\[1em] = \dfrac{\dfrac{1}{\cancel{sin C}} \times \cancel{sin C} - \dfrac{1}{\cancel{tan C}} \times \cancel{tan C}}{\text{sin B}}\\[1em] = \dfrac{1 - 1}{\text{sin B}}\\[1em] = 0$

Hence, $\dfrac{\text{sec A . sin C - tan A . tan C}}{\text{sin B}} = 0$.