Question

The minimum value of ${\sec }^2\alpha +{\cos }^2\alpha$ is

• A. $1$
• B. $2$
• C. $0$
• D. $-1$

Answer 1

Answer: (B)

$2$

$(\sec \alpha -\cos \alpha ){}^2\ge 0$
which implies that
${\sec }^2\alpha +{\cos }^2\alpha -2\ge 0$
${\sec }^2\alpha +{\cos }^2\alpha \ge 2$
Hence minimum value is $2$
Hence answer is B