The smallest tesult of
Attachments:
Answers
Answered by
1
Answer:
Step-by-step explanation:
plz mark it as brainiest
Attachments:
Answered by
1
Answer:
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
Similar questions