Let $f$ be a continuous function going from $[0, \pi]$ to $(-\pi/2,\pi/2)$. I know that $f$ is differentiable and $f(0)=f(\pi)=0$. Prove that there exists $c$ in $(0,\pi)$ so that $f'(c)=\tan(f(c))$.
I have difficulty in finding $g$ that $g'(x)=f'(x)-\tan(f(x))$ to use the Rolle's theorem. The problem is the function $\tan(x)$ is linked with $f(x)$. Write $g(x)=f(x)+\ln(\cos(f(x))$ is not valid. Help me find the function $g$ and the way to find it.
One can apply Rolle's theorem to the function $$ g(x) = e^{-x} \sin(f(x)) \, . $$ $g$ is differentiable with $g(0) = g(\pi) = 0$, so that $g'(c) = 0$ for some $c \in (0, \pi)$. Since $$ g'(x) = e^{-x} \bigl( \cos(f(x)) f'(x) - \sin(f(x))\bigr) $$ and $\cos(f(x)) > 0$, $g'(c) = 0$ is equivalent to $f'(c) = \tan(f(c))$.
Remark: The domain $[0, \pi]$ is irrelevant here, the same proof works for any continuous function $f: [a, b] \to (-\pi/2,\pi/2)$ which is differentiable on $(a, b)$ with $f(a) = f(b) = 0$.