Question

The value of $\frac{3\sin {62}^o}{\cos {28}^o}-\frac{\sec {42}^o}{\text{cosec}{48}^o}$

• A. $2$
• B. $1$
• C. $3$
• D. $0$

Answer 1

The correct option is A $2$

The value of $\frac{3\sin {62}^o}{\cos {28}^o}-\frac{\sec {42}^o}{cosec{48}^o}$ is
$=\frac{3\sin ({90}^o-{28}^o)}{\cos {28}^o}-\frac{\sec ({90}^o-{48}^o)}{cosec{48}^o}$
$=\frac{3\cos {28}^o}{\cos {28}^o}-\frac{cosec{48}^o}{cosec{48}^o}=3-1=2$