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We'll use Lagrange's theorem to prove the trigonometric inequality.
We'll choose a function, whose domain of definition is the closed interval [a,b].
The function is f(x) = tan x
Based on Lagrange's theorem, there is a point "c", that belongs to (a,b), so that:
f(b) - f(a) = f'(c)(b - a)
We'll substitute the function f(x) in the relation above:
tan b - tan a = f'(c)(b-a)
We'll determine f'(x):
f'(x) = 1/(cos x)^2
f'(c) = 1/(cos c)^2
tan b - tan a = (b-a)/(cos c)^2
1/(cos c)^2 = (tan b - tan a)/(b-a)
Since a and b are located in the interval [0 ; pi/2], the cosine function over this interval is decreasing and it has positive values.
a<c<b => cos a > cos c > cos b
We'll raise to square:
(cos a)^2 > (cos c)^2 > (cos b)^2
1/(cos a)^2 < 1/(cos c)^2 < 1/(cos b)^2 (2)
But 1/(cos c)^2 = (tan b - tan a)/(b-a) (1)
From (1) and (2), we'll get:
1/(cos a)^2 < (tan b - tan a)/(b-a) < 1/(cos b)^2
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