f(x) = \left(\frac{u}{\frac{u^2 - 1}{2}}\right)^2 = \left(\frac{2u}{u^2 - 1}\right)^2 = \frac{4u^2}{(u^2 - 1)^2}
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- Let \(u = \sin x + \cos x\). Then:
- u^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x = 1 + 2\sin x \cos x
- \Rightarrow \sin x \cos x = \frac{u^2 - 1}{2}
- Now, since \(0 < x < \frac{\pi}{2}\), \(u = \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)\), so \(u \in (1, \sqrt{2}]\) (maximum \(\sqrt{2}\) at \(x = \frac{\pi}{4}\)).
- Let \(v = u^2\), so \(v \in (1, 2]\). Then:
- f(x) = \frac{4v}{(v - 1)^2}
