Solution: Let $g(x) = f(x) \cos x$. Since \[ g\left( -\frac{\pi}{2} \right) = 0 = g(0) = g\left( \frac{\pi}{2} \right), \] by Rolle's theorem there exists $c_1 \in \left( -\frac{\pi}{2},0 \right) $ and $c_2 \in \left( 0, \frac{\pi}{2} \right) $ such that \[ f^\prime \left( c_1 \right) = f^\prime \left( c_2 \right) = 0. \] Now we consider the function \[ h(x) = \frac{g^\prime (x)}{\cos ^2x} = \frac{f^\prime (x) \cos x - f(x) \sin x}{\cos ^2x}. \]

Since $g(c_1) = g(c_2) =0$, so $h(c_1) = h(c_2) = 0$ and hence again by Rolle's theorem we can find $c \in (c_1, c_2)$ such that $h^\prime \left( c \right) = 0$. That is, \begin{align*} 0 = h^\prime (c) & = \frac{g^{\prime\prime} (c) \cos ^2c - g^\prime (c) 2 \cos c (-\sin c)}{\cos ^4 c} \\[1ex] & = \frac{g^{\prime\prime} (c) \cos c + 2 g^\prime (c) \sin c}{\cos ^3 c} \\[1ex] & = \frac{\left( f^{\prime\prime} (c) \cos c - 2f^\prime (c) \sin c - f(c) \cos c \right) \cos c }{\cos ^3c} \\ & \kern 0.4cm + \frac{2\left( f^\prime (c) \cos c - f(c) \sin c \right) \sin c }{\cos ^3c} \\[1ex] & = \frac{f^{\prime\prime} (c) \cos ^2 c - f(c) \left( \cos ^2 c + 2\sin ^2c \right) }{\cos ^3c} \\[1ex] & = \frac{1}{\cos c} \left( f^{\prime\prime} (c) - f(c) (1+2\tan ^2 c) \right) . \end{align*} Thus, there exists $c \in \left( -\frac{\pi}{2}, \frac{\pi}{2}\right) $ such that \[ f^{\prime\prime} (c) = f(c) \left( 1 + 2\tan ^2c \right) . \]