10. Find the least value of Cos’x+Sec x.
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Answer:
=> 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+se
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