IMC 2013 · Day 1 · P2

Let $g(x) = f(x) \cos x$. Since $g(-\pi/2) = g(0) = g(\pi/2) = 0$, by Rolle's theorem there exist some $\xi_1 \in (-\pi/2, 0)$ and $\xi_2 \in (0, \pi/2)$ such that $$g'(\xi_1) = g'(\xi_2) = 0.$$ Now consider the function $$h(x) = \frac{g'(x)}{\cos^2 x} = \frac{f'(x) \cos x - f(x) \sin x}{\cos^2 x}.$$ We have $h(\xi_1) = h(\xi_2) = 0$, so by Rolle's theorem there exist $\xi \in (\xi_1, \xi_2)$ for which $$\begin{align*} 0 = h'(\xi) &= \frac{g''(\xi) \cos^2 \xi + 2 \cos\xi \sin\xi\, g'(\xi)}{\cos^4 \xi} = \\ &= \frac{(f''(\xi) \cos\xi - 2 f'(\xi) \sin\xi - f(\xi) \cos\xi) \cos\xi + 2 \sin\xi (f'(\xi) \cos\xi - f(\xi) \sin\xi)} {\cos^3 \xi} = \\ &= \frac{f''(\xi) \cos^2 \xi - f(\xi)(\cos^2 \xi + 2 \sin^2 \xi)}{\cos^3 \xi} = \frac{1}{\cos\xi} \bigl( f''(\xi) - f(\xi)(1 + 2 \tan^2 \xi) \bigr). \end{align*}$$ The last yields the desired equality.