Question

If $x={\sec }^2\theta +{\cos }^2\theta$, $\theta \ne 0$ then the value of x is

• A. 0
• B. $\le 2$
• C. $\ge -2$
• D. $\ne 2$

Answer 1

The correct option is D $\ne 2$

${(\sec \theta -\cos \theta )}^2>0$
$\Rightarrow {\sec }^2\theta +{\cos }^2\theta -2\sec \theta \cos \theta >0$
$\Rightarrow {\sec }^2\theta +{\cos }^2\theta -2>0$
$\Rightarrow {\sec }^2\theta +\cos 2\theta -2>0$ i.e $\ne 2$