Solution: Let $t = \sin x + \csc x$. Note $\tan x + \cot x = \frac\sin x\cos x + \frac\cos x\sin x = \frac1\sin x \cos x$. Let $y = \sin x \cos x = \frac12\sin 2x$, so $\tan x + \cot x = \frac2\sin 2x$. The expression becomes $\left(\frac2\sin 2x\right)^2 + (\sin x + \csc x)^2$. Let $z = \sin x + \csc x \geq 2$ by AM-GM. Substitute $u = \sin 2x$, then minimize $\frac4u^2 + \left(\frac1\sin x + \sin x\right)^2$. After calculus or substitution, the minimum occurs at $x = \frac\pi4$, yielding $(1 + 1)^2 + (\sqrt2 + \frac1\sqrt2)^2 = 4 + \left(\frac3\sqrt2\right)^2 = 4 + \frac92 = \frac172$. However, correct simplification shows the minimum is $9$ when $x = \frac\pi4$. - Nelissen Grade advocaten