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# A beautiful equation

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 :)

Note by Nihar Mahajan
1 year, 7 months ago

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

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

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

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. · 1 year, 7 months ago

@Nihar Mahajan How is my solution? Is there any other method u r thinking? · 1 year, 7 months ago

It is quite shorter and elegant than mine.Nice work! · 1 year, 7 months ago

Wait isn't that what I had told you? · 1 year, 7 months ago

That day , we both forgot that $$\cos(A+B+C)=-1$$. Don't worry , it happens sometimes :P · 1 year, 7 months ago

Do u have another method? Can u please post it? · 1 year, 7 months ago

I have another method , but it does not use basic results. It uses "lemma type" results , thats because I found this equation while proving that lemma only! · 1 year, 7 months ago

Please send that solution . I am eager to know another solution to it. Please · 1 year, 7 months ago

Hello Nihar! Can you please tell me from where are you learning trigonometry?? Thanks 😀 · 1 year, 7 months ago

M Prakash Academy , Pune. · 1 year, 7 months ago

Exactly what is "$$cyc$$" in $$\displaystyle \sum_{cyc}$$ and $$\displaystyle\prod_{cyc}$$? $$~~~$$ Never seen this. · 1 year, 7 months ago

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

It is short form of the cyclic summation/product.I hope you get it now :) · 1 year, 7 months ago