Let \(x \to 0^+\), \(\sec x \to 1\), \(\csc x \to \infty\), so sum \(\to \infty\). Similarly near \(\frac{\pi}{2}^-\). So the expression goes to infinity.

February 22, 2026

$Let \(x \to 0^+\), \(\sec x \to 1\), \(\csc x \to \infty\), so sum \(\to \infty\). Similarly near \(\frac{\pi}{2}^-\). So the expression goes to infinity.$

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But we seek the maximum value. Since \(g(v)\) decreases from near infinity (as \(v \to 1^+\)) to 8 at \(v = 2\), but \(v = 1\) is not included. However, as \(x \to 0^+\) or \(\frac{\pi}{2}^-\), one of \(\sin x\) or \(\cos x \to 0\), so \(\sec x\) or \(\csc x \to \infty\), so \(f(x) \to \infty\).
Thus, \((\sec x + \csc x)^2\) is unbounded on \(0 < x < \frac{\pi}{2}\), and has no maximum.
But wait — is that correct?
Hence, there is **no maximum value**; it spans \((8, \infty)\).
But let’s verify at \(x = \frac{\pi}{4}\):
\(\sec \frac{\pi}{4} = \sqrt{2}\), \(\csc \frac{\pi}{4} = \sqrt{2}\), sum = \(2\sqrt{2} \approx 2.828\), square ≈ 8, matches \(g(2) = 8\).

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