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📅 April 16, 2026 👤 scraface

Apr 16, 2026

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Now sum them:
\sec^2 x + \csc^2 x + 2\sec x \csc x + 1 + 2\sin x \cos x
Recall that $\sec^2 x = 1 + \tan^2 x$, $\csc^2 x = 1 + \cot^2 x$, but...
\sec^2 x + \csc^2 x = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} = \frac{...
Also, $\sec x \csc x = \frac{1}{\sin x \cos x}$, so:
2\sec x \csc x = \frac{2}{\sin x \cos x}
And $2\sin x \cos x$ remains.
Let $u = \sin x \cos x$. Since $x \in (0, \frac{\pi}{2})$, $u > 0$, an...
Now rewrite the entire expression in terms of $u$:
\frac{1}{u^2} + \frac{2}{u} + 1 + 2u

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