Question

If $y={cos}^2\theta +{sec}^2\theta ,\theta \ne 0$ then

• A. $y=1$
• B. $y\underset{–}{<}2$
• C. $y\underset{–}{>}-2$
• D. $y\underset{–}{>}2$

Answer 1

The correct option is D $y\underset{–}{>}2$

Given,

$y={\cos }^2\theta +{\sec }^2\theta$

$⟹y={\sec }^2\theta -2\sec \theta \cos \theta +{\cos }^2\theta +2\sec \theta \cos \theta$

$⟹y=(\sec \theta -\cos \theta {)}^2+2$

As $2\sec \theta \cos \theta =2$

Sl$⟹(\sec \theta -\cos \theta {)}^2\ge 0$

$⟹(\sec \theta -\cos \theta {)}^2+2\ge 0+2$

$\therefore y={\cos }^2\theta +{\sec }^2\theta \ge 2⟹y\ge 2.$