Question

solve this please.. ......

Answer 1

Answer:

Option(b)

Step-by-step explanation:

Given: sec²θ + cos²θ

(i) A.M of the given equation = (sec²θ + cos²θ)/2

(i) G.M of the given equation = √sec²θ * cos²θ

Now, we know that A.M ≥ G.M, we get

⇒ (sec²θ + cos²θ)/2 ≥ (√sec²θ * cos²θ)

⇒ (sec²θ + cos²θ)/2 ≥ 1

⇒ sec²θ + cos²θ ≥ 2

∴ Lowest value is 2

Hope it helps!

Answer 2

Step-by-step explanation:

LET a = (sec^2)x , b = (cos^2)x , so a, b >0

a+b ≥ 2√ ab) ,

AM ≥GM , (sec^2)x +(cos^2)x ≥ 2 √ ( (sec^2)x* (cos^2)x ≥ 2

Least value is 2 and max tends to infinity