You need to find out if the mean value theorem can be applied to the the given function, hence you need to verify if the function is continuous on [0,pi] and differentiable on (0,pi), and it is, since all trigonometric functions are continuous and differentiable on the interval.
Since the mean value theorem can be applied, then there is a point c in (0,pi), such that:
f(pi) - f(0) = f'(c)(pi-0)
You need to evaluate f(pi) and f(0):
f(pi) = cos pi + tan pi => f(pi) =-1 + 0 = -1
f(0) = cos 0 + tan 0 => f(0) = 1 + 0 = 1
You need to determine f'(c):
f'(c) = -sin c + 1/(cos^2 c)
Replacing the found values in equation f(pi) - f(0) = f'(c)(pi-0) yields:
-1-1 = pi*(1/(cos^2 c) - sin c)
-2 = pi*(1/(cos^2 c) - sin c)
Replace 1 - sin^2 c for cos^2 c :
-2 = pi*(1/(1 - sin^2 c) - sin c)
You need to use the substitution sin c = v:
-2 = pi*(1/(1 - v^2) - v) => -2 = pi(1 - v + v^3)/(1 - v^2)
2v^2 - 2 = pi - pi*v + pi*v^3
pi*v^3 - 2v^2 - pi*v + pi + 2 = 0
Since there is no solution to the given equation pi*v^3 - 2v^2 - pi*v + pi + 2 = 0 , then there is no valid value of c in (0,pi), such as f'(c) = 0 and the mean value theorem cannot be applied.
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