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This question is in error. (The source had a typo. See this answer.)

Let $a$ be a real positive number. \begin{align} \text{(I)}& & a &> \frac{\sin(y_1(a))}{y_1(a)} & &\text{where $y_1(a)$ is the unique root of}& y &= a \cot(y) ~~\text{in}~~ (0,\frac{\pi}{2}) \\ \text{(II)}& & a &>\xi & &\text{where $~~~\xi~\,~~$ is the unique root of}& \xi^2 &= \cos(\xi) \quad\text{in}~~ (0,\frac{\pi}{2}) \end{align}

I have read that the two statements are equivalent, i.e. (I)$\iff$(II). Does anyone have any hint how to prove this equivalence?

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    $\begingroup$ Is there a typo? I think for inequality (I) the defining equation for $y_1(a)$ should be $y = a \tan y$. That way things work out trivially. $\endgroup$ – Lee David Chung Lin Apr 21 at 11:55
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I have eventually found out there was a typo in the book where I read this claim. The correct statement which can be proved is the following.

Let $a$ be a real positive number. \begin{align} \text{(I)}& & a &> \frac{\sin(y_1(a))}{y_1(a)} & &\text{where $y_1(a)$ is the unique root of}& y &= a \cot(y) ~~\text{in}~~ (0,\frac{\pi}{2}) \\ \text{(II)}& & a &>\frac{\sin\xi}{\xi} & &\text{where $~~~\xi~\,~~$ is the unique root of}& \xi^2 &= \cos(\xi) \quad\text{in}~~ (0,\frac{\pi}{2}) \end{align}

The statement in the question is not true.

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