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What is the range of $ y = (\operatorname{ arccot x }) (\operatorname{ arccot{ - x }}) $. I solved this problem with right answer using AM GM inequality. But I received a lot of criticism for using AM GM inequality here on this site as it does not give sharp bounds. So is there a better way? I was thinking about Jensen's inequality but that doesn't work.

What is wrong with my solution of maximum value of $ \sin \frac {A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} $ in a triangle ABC?

The side of a triangle inscribed in a given circle subtends angles $a, b, $ and $ y$ at the center.

What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$

user541396
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    Can you please provide some context and show what you have tried? What is more, there are some issues with your LaTeX code. Use \operatorname{arccot} instead of \arccot. – Pantelis Sopasakis May 14 '19 at 12:52
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    @PantelisSopasakis I did say I solved this problem with right answer using AM GM what more could I add? – user541396 May 14 '19 at 12:54
  • You could read this guide on writing a good question. You could offer people some context, show them what you've tried. You say you received criticism of MSE for using AM/GM; can you provide a link? BTW: a better way to typeset your equation: y = (\operatorname{arccot} x) (\operatorname{arccot}(-x)) – Pantelis Sopasakis May 14 '19 at 12:57
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    @PantelisSopasakis anything else I should add?? – user541396 May 14 '19 at 13:03
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    @PantelisSopasakis questions are put on hold even if they follow "this guide"https://math.meta.stackexchange.com/questions/30088/why-was-my-question-put-on-hold. So I doubt how helpful it is. – user541396 May 14 '19 at 13:14
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    You're right, MSE doesn't work perfectly. I meant to say that the better you phrase your questions, the higher the chances that you'll get an answer. – Pantelis Sopasakis May 14 '19 at 13:18

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Like How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?,

arccot$(x)\cdot$arccot$(-x)=$arccot$(x)(\pi-$arccot$(x))=\left(\dfrac\pi2\right)^2-\left(\text{arccot }(x)-\dfrac\pi2\right)^2$

lab bhattacharjee
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