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How can I prove that for all $t\in[0,\frac{\pi}{2}], \cot^2t\leq\frac{1}{t^2}\leq1+\cot^2t$, with $\cot$ the cotangent function ? Thank you

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Are you sure you have written the inequality right? – Fredrik Meyer Jan 11 '12 at 11:20
Pretty much, yes. Why ? – user20010 Jan 11 '12 at 11:28
I think we can derive a simpler inequality to prove from the given inequality, and prove it by derivatives or similar. – user20010 Jan 11 '12 at 11:29

Start with $0<\sin x < x < \tan x$, square, and take reciprocals.

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See also…. – Hans Lundmark Jan 11 '12 at 11:30

This is an elaboration of Hans Lundmark's answer:

Consider the portion of the unit circle in the first quadrant. For an angle $t$ in the first quadrant, consider the triangle $\triangle OIQ$ where $O=(0,0)$, $I=(1,0)$, and $Q=(\cos t, \sin t)$. Let $Z$ be the point of intersection of the line passing through $O$ and $Q$ with the vertical line passing through $I$.

Then $$ \text{area}(\triangle OIQ)\le\text{area}( \text{sector}(OIQ))\le\text{area}(\triangle OIZ). $$

This gives $$ {1\over2}\sin t\le {t\over 2}\le {1\over2}\tan t; $$ whence $$ {1\over \sin t}\ge{1\over t}\ge {\cos t\over \sin t}. $$ From this $$\eqalign{ &\cot t \le{1\over t}\le \csc t\cr \iff& \cot^2 t\le {1\over t^2}\le \csc^2 t\cr \iff&\cot^2\le {1\over t^2}\le 1+\cot^2 t, } $$ where we used the Pythagorean identity $1+\cot^2 t=\csc^2 t$ in the last inequality.

enter image description here

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I'd like to present a Calculus approach to the question, even if I don't know if it can satisfy the OP (because of the restriction imposed by the Trigonometry tag).

Let $x=\tan t$, thus the inequality becomes: $$\frac{1}{x^2} \leq \frac{1}{\arctan^2 x} \leq 1+\frac{1}{x^2} =\frac{x^2+1}{x^2}$$ with $x\in ]0,\infty[$; therefore, taking the reciprocals, it suffices to prove: $$\frac{x^2}{x^2+1} \leq \arctan^2 x \leq x^2\; .$$ The rightmost inequality is trivial, for we know that $\arctan x\leq x$ when $x\geq 0$, hence we have to work on the leftmost inequality.

Now, taking the squares the inequality to prove becomes: $$\frac{x}{\sqrt{1+x^2}} \leq \arctan x \; ,$$ or, equivalently: $$\tag{1} \arctan x -\frac{x}{\sqrt{1+x^2}} \geq 0\; .$$ Let, for sake of simplicity, $f(x):=\arctan x-\frac{x}{\sqrt{1+x^2}}$: since $f(0)=0$, we can prove inequality (1) by showing that $f$ increases in $[0,\infty[$. In order to do this, we evaluate the derivative of $f$: $$f^\prime (x) = \frac{1}{1+x^2} -\frac{1}{(1+x^2)\sqrt{1+x^2}}=\frac{\sqrt{1+x^2} -1}{(1+x^2)\sqrt{1+x^2}}$$ and we see at once that $f^\prime (x)\geq 0$ when $x\geq 0$ (because $\sqrt{1+x^2}\geq 1$) thus $f$ increases strictly in $[0,\infty[$ and $f(x)\geq f(0)$, which is (1).

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