Prove that sec^2x+cos^2x can never be less than 2
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Answered by
37
Since, Arithmatic mean ≥ Geometric Mean
Then for any x,y we have,
(x+y) ≥ 2√xy
Let, x=a and y=1/a
∴, (a+1/a) ≥ 2√(a×1/a)
or, a+1/a ≥ 2
Putting, a=cos²x we get,
cos²x+1/cos²x ≥ 2
or, cos²x+sec²x ≥ 2 (Proved)
Then for any x,y we have,
(x+y) ≥ 2√xy
Let, x=a and y=1/a
∴, (a+1/a) ≥ 2√(a×1/a)
or, a+1/a ≥ 2
Putting, a=cos²x we get,
cos²x+1/cos²x ≥ 2
or, cos²x+sec²x ≥ 2 (Proved)
Answered by
22
Derivatives method:
let cos x = y , sec x = 1/y
Let M = y^2 + 1/y^2
dM/dy = 2 y - 2/y^3 = 0 => y = +1 or -1
d^2 M/dy^2 = 2 + 6 / y^4 > 0
so at y = 1 or -1, (at x=0) , we have a minimum value for M.
Minimum value of M = 1 + 1 = 2
let cos x = y , sec x = 1/y
Let M = y^2 + 1/y^2
dM/dy = 2 y - 2/y^3 = 0 => y = +1 or -1
d^2 M/dy^2 = 2 + 6 / y^4 > 0
so at y = 1 or -1, (at x=0) , we have a minimum value for M.
Minimum value of M = 1 + 1 = 2
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