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- \sec^2 x = 1 + \tan^2 x,\quad \csc^2 x = 1 + \cot^2 x
- Now, note that:
- \tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}
- Let $ s = \sin x + \cos x $, $ p = \sin x \cos x $. Then:
- s^2 = 1 + 2p \Rightarrow p = \frac{s^2 - 1}{2}
- We minimize this expression for $ s \in (\sqrt{2}, \sqrt{2}] $, since $ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) $, and maximum is $ \sqrt{2} $. Wait — maximum is $ \sqrt{2} $, minimum is 1 (approaching at endpoints), but in the open interval $ (0, \frac{\pi}{2}) $, $ s \in (1, \sqrt{2}) $.
