Q:

Minimum value of Function

I have a function, $f(x) = \sec x + \csc x$ on the interval $x \in (0, \pi/2)$.
The derivative of it is, $f'(x) = \csc^2(x) \sec^2(x) \left(\sin^3 x + \cos^3 x\right)$
Of course, when I tried to solve for $f'(x) = 0$, I realize that $f'(x)=0$ does not exist. 
However, the question insists that the there is a minimum. Am I doing something wrong here?

A:

We should have $$\begin{align}f'(x) &= \frac{\sin x}{\cos^2 x}-\frac{\cos x}{\sin^2x}\\ &= \frac1{\cos^2x\sin^2x}\bigl(\sin^3x-\cos^3x\bigr)\\ &= \frac1{\cos^2x\sin^2x}(\sin x-\cos x)\bigl(\sin^2x+\sin x\cos x+\cos^2x\bigr)\\ &= \frac1{\cos^2x\sin^2x}(\sin x-\cos x)\left(1+\frac12\sin 2x\right).\end{align}$$ The fraction and the right-most factor at the end can't be zero for any real $x$. Hence, $f'(x)=0$ if and only if $\sin x=\cos x$. Where does that happen?

