Question Number 22536 by NECx last updated on 20/Oct/17
$${show}\:{that}\:\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }= \\ $$$$\frac{\mathrm{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{A}+\mathrm{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{B}+\mathrm{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{C}}{\mathrm{cot}\:{A}+\mathrm{cot}\:{B}+\mathrm{cot}\:{C}} \\ $$$$ \\ $$$$ \\ $$$${please}\:{help} \\ $$
Commented by ajfour last updated on 20/Oct/17
$${Another}\:{solution}:\:{see}\:{Q}.\mathrm{22576} \\ $$
Answered by $@ty@m last updated on 20/Oct/17
$${We}\:{have} \\ $$$$\mathrm{cot}\:\frac{{A}}{\mathrm{2}}=\sqrt{\frac{{s}\left({s}−{a}\right)}{\left({s}−{b}\right)\left({s}−{c}\right)}}\:\:\:−−\left(\mathrm{1}\right) \\ $$$${Let}\:\frac{\mathrm{sin}\:{A}}{{a}}=\frac{\mathrm{sin}\:{B}}{{b}}=\frac{\mathrm{sin}\:{C}}{{c}}={k} \\ $$$$\Rightarrow\mathrm{cot}\:{A}=\frac{\mathrm{cos}\:{A}}{\mathrm{sin}\:{A}}=\frac{\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}}{{ak}}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{abck}}\:\:−−\left(\mathrm{2}\right) \\ $$$${Using}\:\left(\mathrm{1}\right)\:\&\:\left(\mathrm{2}\right) \\ $$$${R}.{H}.{S}.=\frac{\mathrm{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{A}+\mathrm{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{B}+\mathrm{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{C}}{\mathrm{cot}\:{A}+\mathrm{cot}\:{B}+\mathrm{cot}\:{C}} \\ $$$$=\frac{\sqrt{\frac{{s}\left({s}−{a}\right)}{\left({s}−{b}\right)\left({s}−{c}\right)}}+\sqrt{\frac{{s}\left({s}−{b}\right)}{\left({s}−{a}\right)\left({s}−{c}\right)}}+\sqrt{\frac{{s}\left({s}−{c}\right)}{\left({s}−{b}\right)\left({s}−{a}\right)}}}{\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{abck}}+\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{abck}}+\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{abck}}} \\ $$$$=\frac{\mathrm{2}{abck}}{\:\sqrt{\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}}.\frac{\sqrt{{s}\left({s}−{a}\right)^{\mathrm{2}} }+\sqrt{{s}\left({s}−{b}\right)^{\mathrm{2}} }+\sqrt{{s}\left({s}−{c}\right)^{\mathrm{2}} }}{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} +{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} +{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{abck}}{\:\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}}.\frac{{s}\left\{\left({s}−{a}\right)+\left({s}−{b}\right)+\left({s}−{c}\right)\right\}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{abck}}{\frac{\mathrm{1}}{\mathrm{2}}{bc}\mathrm{sin}\:{A}}.\frac{{s}\left\{\mathrm{3}{s}−\left({a}+{b}+{c}\right)\right\}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$=\mathrm{4}.\frac{{s}\left\{\mathrm{3}{s}−\mathrm{2}{s}\right\}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$=\mathrm{4}.\frac{{s}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$=\frac{\left(\mathrm{2}{s}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$=\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }={L}.{H}.{S}. \\ $$
Commented by NECx last updated on 20/Oct/17
$${wow}…..\:{i}'{m}\:{most}\:{grateful}. \\ $$