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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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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