Question

find the least value of cos²x+sec²x
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Answer 1

Here, we have to find out the least value of f(x)=cos²x+sec²x

Now,

{cos(x) - 1/cos(x)}² ≥ 0

(Since, all the squares are always positive/non-negative.)

=> cos²x + 1/cos²x - 2 × cos(x) × 1/cos(x) ≥ 0

=> cos²x + sec²x - 2 ≥ 0

{Since, sec(x) is the reciprocal of cos(x)}

=> cos²x + sec²x ≥ 2

Therefore, f(x)=cos²x+sec²x ∈ (2,∞)

Hence, the least value of f(x)=cos²x+sec²x is “2”.

Hope, it helps. <3

cheers ✌️

Answer 2

Answer:

cos²x+sec²x

1/sec²x+sec²x

1+sec^4x /sec²x