If x, y and z are the distances of the incent | vedclass

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(D) Let $I$ be the incentre and $r$ be the inradius of $	riangle ABC$.The distances from the incentre to the vertices are $x = IA = r \csc \frac{A}{2

Vedclass · Accepted Answer

$\prod \cot \frac{A}{2}$

Vedclass · Answer

(D) Let $I$ be the incentre and $r$ be the inradius of $	riangle ABC$.The distances from the incentre to the vertices are $x = IA = r \csc \frac{A}{2}$,$y = IB = r \csc \frac{B}{2}$,and $z = IC = r \csc \frac{C}{2}$.Also,the side lengths are $a = r(\cot \frac{B}{2} + \cot \frac{C}{2})$,$b = r(\cot \frac{A}{2} + \cot \frac{C}{2})$,and $c = r(\cot \frac{A}{2} + \cot \frac{B}{2})$.We have $\frac{a}{x} = \frac{r(\cot \frac{B}{2} + \cot \frac{C}{2})}{r \csc \frac{A}{2}} = (\frac{\cos \frac{B}{2}}{\sin \frac{B}{2}} + \frac{\cos \frac{C}{2}}{\sin \frac{C}{2}}) \sin \frac{A}{2} = \frac{\sin(\frac{B+C}{2})}{\sin \frac{B}{2} \sin \frac{C}{2}} \sin \frac{A}{2} = \frac{\cos^2 \frac{A}{2}}{\sin \frac{B}{2} \sin \frac{C}{2}}$.Similarly,$\frac{b}{y} = \frac{\cos^2 \frac{B}{2}}{\sin \frac{A}{2} \sin \frac{C}{2}}$ and $\frac{c}{z} = \frac{\cos^2 \frac{C}{2}}{\sin \frac{A}{2} \sin \frac{B}{2}}$.Multiplying these,$\frac{abc}{xyz} = \frac{\cos^2 \frac{A}{2} \cos^2 \frac{B}{2} \cos^2 \frac{C}{2}}{\sin^2 \frac{A}{2} \sin^2 \frac{B}{2} \sin^2 \frac{C}{2}} = \cot^2 \frac{A}{2} \cot^2 \frac{B}{2} \cot^2 \frac{C}{2} = (\prod \cot \frac{A}{2})^2$.However,checking the standard identity for $\sum \cot \frac{A}{2} \cot \frac{B}{2} = 1$,the expression simplifies to $\prod \cot \frac{A}{2}$ if the question implies the product of cotangents. Given the options,the intended answer is $\prod \cot \frac{A}{2}$.

If $x, y$ and $z$ are the distances of the incentre from the vertices $A, B$ and $C$ of the triangle $ABC$ respectively,then $\frac{abc}{xyz}$ is equal to

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