Indeed the saying "Curiosity is the mother of inventions" is absolutely true. While learning trigonometry , I found this:

Prove that in $\Delta ABC$ ,

$\large\sum_{cyc} \cot A = \prod_{cyc} \csc A + \prod_{cyc} \cot A$

I invite all brilliantians to prove this :)

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## Comments

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TopNewestSince, $A+B+C = 180^{o}$

$Cos(A+B+C) = -1$

$\prod_{cyc} \cos (A) - \sum_{cyc} \cos (A) \sin (B) \sin (C) = -1$

Divide both sides with $\prod_{cyc} \sin (A)$,

$\prod_{cyc} \cot (A) - \sum_{cyc} \cot (A) = - \prod_{cyc} \csc (A)$

$\sum_{cyc} \cot (A) = \prod_{cyc} \cot (A) + \prod_{cyc} \csc (A)$

Hence, proved.

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@Nihar Mahajan How is my solution? Is there any other method u r thinking?

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It is quite shorter and elegant than mine.Nice work!

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$\cos(A+B+C)=-1$. Don't worry , it happens sometimes :P

That day , we both forgot thatLog in to reply

Hello Nihar! Can you please tell me from where are you learning trigonometry?? Thanks 😀

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M Prakash Academy , Pune.

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Exactly what is "$cyc$" in $\displaystyle \sum_{cyc}$ and $\displaystyle\prod_{cyc}$? $~~~$ Never seen this.

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$\sum_{cyc} \cot A = \cot A+\cot B+\cot C \\ \prod_{cyc} \cot A = \cot A\cot B\cot C$

It is short form of the cyclic summation/product.I hope you get it now :)

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what does sym summation mean?

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