﻿ For any triangle ABC, prove that cot (A/2)+cot(B/2)+cot(C/2) = cot(A/2)cot(B/2)cot(C/2) \cot\dfrac{A}{2}+ \cot\dfrac{B}{2}+\cot\dfrac{C}{2} = \cot\dfrac{A}{2}\cot\dfrac{B}{2}\cot\dfrac{C}{2}

#### Question

For any triangle ABC, prove that cot (A/2)+cot(B/2)+cot(C/2) = cot(A/2)cot(B/2)cot(C/2)

\cot\dfrac{A}{2}+ \cot\dfrac{B}{2}+\cot\dfrac{C}{2} = \cot\dfrac{A}{2}\cot\dfrac{B}{2}\cot\dfrac{C}{2}

Collected in the board: Trigonometry

Steven Zheng posted 1 day ago

For the three internal angles of a triangle

A+B+C=\pi

Then,

\cot(\dfrac{A}{2}+ \dfrac{B}{2} )=\cot( \dfrac{\pi}{2}-\dfrac{C}{2})

Apply the sum identity for cotangent function on both sides.

The cotangent of the sum of two angles equals the product of the cotangents of the angles minus 1 divided by the sum of the cotangents of the two angles

Considering \cot\dfrac{-C}{2}= -\cot\dfrac{C}{2}, we get

\dfrac{\cot \dfrac{A}{2}\cot \dfrac{B}{2} -1 }{\cot \dfrac{A}{2}+\cot \dfrac{B}{2} } = \dfrac{-\cot \dfrac{\pi}{2}\cot \dfrac{C}{2}-1 }{\cot \dfrac{\pi}{2}-\cot \dfrac{C}{2}}

Since \cot\dfrac{\pi}{2}=0, the equation is simplified to

\dfrac{\cot \dfrac{A}{2}\cot \dfrac{B}{2} -1 }{\cot \dfrac{A}{2}+\cot \dfrac{B}{2} } = \dfrac{1}{\cot \dfrac{C}{2}}

Cross-Multiply of the equation gives

\cot\dfrac{A}{2}\cot\dfrac{B}{2}\cot\dfrac{C}{2}-1=\cot\dfrac{A}{2}+ \cot\dfrac{B}{2}

Hence

\cot\dfrac{A}{2}+ \cot\dfrac{B}{2}+\cot\dfrac{C}{2} = \cot\dfrac{A}{2}\cot\dfrac{B}{2}\cot\dfrac{C}{2}

Steven Zheng posted 1 day ago

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