Question

pls tell the answer past​

Answer 1

Step-by-step explanation:

$\because \: \triangle ABC \: is \: right \: angled \: at \: B \\ \therefore \angle \: B = 90 \degree \\ \therefore \: A + B + C = 180 \degree \\ \\ \therefore \: A + 90 \degree + C = 180 \degree \\ \\ \therefore \: A + C = 90 \degree \\ \\ \therefore \: A = 90 \degree - C \\ \\ \therefore \: \frac{sec A \: sin C - tan A \: tan C}{sin B} \\ \\ = \frac{sec ( 90 \degree - C ) \: sin C - tan ( 90 \degree - C ) \: tan C}{sin 90 \degree} \\ \\ = \frac{cosec C \: sin C - cot C\: tan C}{1} \\ \\ = \frac{1}{sin C} \times sin C - \frac{1}{tan C} \times tan C \\ \\ = 1 - 1 \\ \\ = 0 \\ \\ thus \\ \\ \purple{\boxed{ \therefore \: \frac{sec A \: sin C - tan A \: tan C}{sin B} = 0}}$