If x,y and z are the distances of incenter from the vertices of the triangle ABC, respectively, then prove that $\frac{a b c}{x y z} =$

cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2}

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tan \frac{A}{2} tan \frac{B}{2} tan \frac{C}{2}

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cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}

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sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

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Solution

The correct option is A $cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2}$
$x = r c o s e c \frac{A}{2}$ and $a = r (cot \frac{B}{2} + cot \frac{C}{2})$ $\Rightarrow \frac{a}{x} = (cot \frac{B}{2} + cot \frac{C}{2}) sin \frac{A}{2} = \frac{sin \frac{A}{2} cos \frac{A}{2}}{sin \frac{B}{2} sin \frac{C}{2}}$ or
$\frac{a b c}{x y z} = \frac{cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}}{sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}} = cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2}$

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A, B

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cos A + cos B + cos C - 1

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∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 cot \frac{A}{2} & cot \frac{B}{2} & cot \frac{C}{2} tan \frac{B}{2} + tan \frac{C}{2} & tan \frac{C}{2} + tan \frac{A}{2} & tan \frac{A}{2} + tan \frac{B}{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 0

In a triangle ABC,
cosA- cosB=

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