If y = cos² x + sec² x, where 0 ≤ x < \(\frac{π}{2}\), then which one of the following is correct?

Given:

y = cos2 x + sec2 x; 0 ≤ x < \(\frac{π}{2}\)

Concept Used:

• A.M ≥ G.M
• If a and b are two numbers then (a + b)/2 ≥ √ab
• cos2 x × sec2 x = 1

Calculation:

We have y = cos2 x + sec2 x     ----(i)

Let the two numbers be cos2 x and sec2 x 

According to the concept used

A.M ≥ G.M

⇒ [cos2 x + sec2 x]/2 ≥ √[cos2 x × sec2 x]

⇒ [cos2 x + sec2 x]/2 ≥ 1

⇒ [cos2 x + sec2 x]  ≥ 2

Now, from (i)

⇒ y ≥ 2

∴ The correct relation is y ≥ 2.