Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 46

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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