In ABC prove that A 2 + B 2 + C 2 =a+b+c b+c-a A 2 - Vidyadip

Question

In $\triangle ABC$ prove that $\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\frac{a+b+c}{b+c-a} \cot \frac{A}{2}$

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Answer

$\text { L.H.S. }=\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}$
$=\frac{1}{\tan \frac{A}{2}}+\frac{1}{\tan \frac{B}{2}}+\frac{1}{\tan \frac{C}{2}}$
$=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}+\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}+\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$
$=\sqrt{\frac{s(s-a)^2}{(s-b)(s-c)(s-a)}}+\sqrt{\frac{s(s-b)^2}{(s-a)(s-c)(s-b)}}+\sqrt{\frac{s(s-c)^2}{(s-b)(s-a)(s-c)}}$
$=\sqrt{\frac{s}{(s-b)(s-a)(s-c)}}\left\{\sqrt{(s-a)^2}+\sqrt{(s-b)^2}+\sqrt{(s-c)^2}\right\}$
$=\sqrt{\frac{s}{(s-b)(s-a)(s-c)}}\{(s-a)+(s-c)+(s-b)\}$
$=\sqrt{\frac{s}{(s-b)(s-a)(s-c)}}\{3 s-(a+b+c)\}$
$=\sqrt{\frac{s}{(s-b)(s-a)(s-c)}}\{3 s-2 s\}$
$=\sqrt{\frac{s}{(s-b)(s-a)(s-c)}} \times s$
$=\sqrt{\frac{s}{(s-b)(s-c)}} \times \frac{s}{\sqrt{(s-a)}}$
$=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}} \times \frac{s}{(s-a)}$
$=\frac{2 s}{(2 s-2 a)} \times \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$
$=\frac{a+b+c}{(a+b+c-2 a)} \times \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$
$=\frac{a+b+c}{b+c-a} \cot \frac{A}{2}=\text { R.H.S. }$

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