Question

If $A$ and $B$ are complementary angles, then $\sin A\times \sec B=$

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

The correct option is A $1$

Given that both the angles are complementary i.e., $A+B={90}^{\circ }$

So, $A={90}^{\circ }-B$

We have :

$=\sin A\times \sec B$

$=\sin ({90}^o-B)\times \sec B$

$=\cos B\times \sec B$ [using $\cos \theta =\sin ({90}^o-\theta )$]

$=\cos B\times \frac{1}{\cos B}$

$=1$