Question

A triangle ABC is right angled at B. Find the value of

$\frac{\sec A.\sin C-\tan A\tan C}{\sin B}$

Answer 1

Since triangle ABC is a right-angled triangle, and right angle at B

Therefore, $B={90}^{\circ },A+C={90}^{\circ }$

$\frac{\sec A.\sin C-\tan A\tan C}{\sin B}$

$=\frac{cosecC.\sin C-\tan (90-C)\tan C}{\sin {90}^{\circ }}$

$=\frac{cosecC.\sin C-\cot C\tan C}{1}$

$=\frac{1}{\sin C}.\sin C-\frac{1}{\tan C}\tan C$

On simplifying we get

$=1-1$

$=0$