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 :)

#Geometry

Easy Math Editor

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Comments

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.

Surya Prakash - 5 years, 10 months ago

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

Surya Prakash - 5 years, 10 months ago

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