But the sum is minimized at $8$ (when $x = \frac{\pi}{4}$), but maximum is unbounded. However, the problem likely seeks the minimum. Assuming a typo, if the question is to find the minimum:

February 22, 2026

$But the sum is minimized at $8$ (when $x = \frac{\pi}{4}$), but maximum is unbounded. However, the problem likely seeks the minimum. Assuming a typo, if the question is to find the minimum:$

+ t + \frac{1}{t}.
By AM-GM, $t + \frac{1}{t} \geq 2$, with equality when $t = 1$ (i.e., $x = \frac{\pi}{4}$). Thus, the minimum is $7$, but we seek the maximum. As $x \to 0^+$ or $x \to \frac{\pi}{2}^-$, $\tan^2 x \to 0$ or $\infty$, so $t + \frac{1}{t} \to \infty$. However, the original expression is unbounded. Wait, this contradicts the problem's implication of a finite maximum. Re-examining:
Actually, $\sin x + \csc x = \sin x + \frac{1}{\sin x} \geq 2$, and similarly for $\cos x + \sec x$. But squaring gives:
At $x = \frac{\pi}{4}$, $\sin x = \cos x = \frac{\sqrt{2}}{2}$, $\csc x = \sec x = \sqrt{2}$. Then:
\left(\frac{\sqrt{2}}{2} + \sqrt{2}\right)^2 + \left(\frac{\sqrt{2}}{2} + \sqrt{2}\right)^2 = 2 \cdot \left(\frac{3\sqrt{2}}{2}\right)^2 = 2 \cdot \frac{18}{4} = 9.
Thus, the minimum is $9$. But the question asks for the maximum, which is unbounded. Clarifying, if the original expression is correct, the maximum is $\infty$. However, assuming a misinterpretation, the intended answer is likely:

But the sum is minimized at $8$ (when $x = \frac{\pi}{4}$), but maximum is unbounded. However, the problem likely seeks the minimum. Assuming a typo, if the question is to find the minimum:

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