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Question Number 22536 by NECx last updated on 20/Oct/17

show that (((a+b+c)^2 )/(a^2 +b^2 +c^2 ))= ((cot (1/2)A+cot (1/2)B+cot (1/2)C)/(cot A+cot B+cot C)) please help

$${show}\:{that}\:\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }= \\ $$$$\frac{\mathrm{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{A}+\mathrm{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{B}+\mathrm{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{C}}{\mathrm{cot}\:{A}+\mathrm{cot}\:{B}+\mathrm{cot}\:{C}} \\ $$$$ \\ $$$$ \\ $$$${please}\:{help} \\ $$

Commented by ajfour last updated on 20/Oct/17

Another solution: see Q.22576

$${Another}\:{solution}:\:{see}\:{Q}.\mathrm{22576} \\ $$

Answered by $@ty@m last updated on 20/Oct/17

We have cot (A/2)=(√((s(s−a))/((s−b)(s−c)))) −−(1) Let ((sin A)/a)=((sin B)/b)=((sin C)/c)=k ⇒cot A=((cos A)/(sin A))=(((b^2 +c^2 −a^2 )/(2bc))/(ak))=((b^2 +c^2 −a^2 )/(2abck)) −−(2) Using (1) & (2) R.H.S.=((cot (1/2)A+cot (1/2)B+cot (1/2)C)/(cot A+cot B+cot C)) =(((√((s(s−a))/((s−b)(s−c))))+(√((s(s−b))/((s−a)(s−c))))+(√((s(s−c))/((s−b)(s−a)))))/(((b^2 +c^2 −a^2 )/(2abck))+((a^2 +c^2 −b^2 )/(2abck))+((a^2 +b^2 −c^2 )/(2abck)))) =((2abck)/(√((s−a)(s−b)(s−c)))).(((√(s(s−a)^2 ))+(√(s(s−b)^2 ))+(√(s(s−c)^2 )))/(b^2 +c^2 −a^2 +a^2 +c^2 −b^2 +a^2 +b^2 −c^2 )) =((2abck)/(√(s(s−a)(s−b)(s−c)))).((s{(s−a)+(s−b)+(s−c)})/(a^2 +b^2 +c^2 )) =((2abck)/((1/2)bcsin A)).((s{3s−(a+b+c)})/(a^2 +b^2 +c^2 )) =4.((s{3s−2s})/(a^2 +b^2 +c^2 )) =4.(s^2 /(a^2 +b^2 +c^2 )) =(((2s)^2 )/(a^2 +b^2 +c^2 )) =(((a+b+c)^2 )/(a^2 +b^2 +c^2 ))=L.H.S.

$${We}\:{have} \\ $$$$\mathrm{cot}\:\frac{{A}}{\mathrm{2}}=\sqrt{\frac{{s}\left({s}−{a}\right)}{\left({s}−{b}\right)\left({s}−{c}\right)}}\:\:\:−−\left(\mathrm{1}\right) \\ $$$${Let}\:\frac{\mathrm{sin}\:{A}}{{a}}=\frac{\mathrm{sin}\:{B}}{{b}}=\frac{\mathrm{sin}\:{C}}{{c}}={k} \\ $$$$\Rightarrow\mathrm{cot}\:{A}=\frac{\mathrm{cos}\:{A}}{\mathrm{sin}\:{A}}=\frac{\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}}{{ak}}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{abck}}\:\:−−\left(\mathrm{2}\right) \\ $$$${Using}\:\left(\mathrm{1}\right)\:\&\:\left(\mathrm{2}\right) \\ $$$${R}.{H}.{S}.=\frac{\mathrm{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{A}+\mathrm{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{B}+\mathrm{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{C}}{\mathrm{cot}\:{A}+\mathrm{cot}\:{B}+\mathrm{cot}\:{C}} \\ $$$$=\frac{\sqrt{\frac{{s}\left({s}−{a}\right)}{\left({s}−{b}\right)\left({s}−{c}\right)}}+\sqrt{\frac{{s}\left({s}−{b}\right)}{\left({s}−{a}\right)\left({s}−{c}\right)}}+\sqrt{\frac{{s}\left({s}−{c}\right)}{\left({s}−{b}\right)\left({s}−{a}\right)}}}{\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{abck}}+\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{abck}}+\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{abck}}} \\ $$$$=\frac{\mathrm{2}{abck}}{\sqrt{\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}}.\frac{\sqrt{{s}\left({s}−{a}\right)^{\mathrm{2}} }+\sqrt{{s}\left({s}−{b}\right)^{\mathrm{2}} }+\sqrt{{s}\left({s}−{c}\right)^{\mathrm{2}} }}{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} +{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} +{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{abck}}{\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}}.\frac{{s}\left\{\left({s}−{a}\right)+\left({s}−{b}\right)+\left({s}−{c}\right)\right\}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{abck}}{\frac{\mathrm{1}}{\mathrm{2}}{bc}\mathrm{sin}\:{A}}.\frac{{s}\left\{\mathrm{3}{s}−\left({a}+{b}+{c}\right)\right\}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$=\mathrm{4}.\frac{{s}\left\{\mathrm{3}{s}−\mathrm{2}{s}\right\}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$=\mathrm{4}.\frac{{s}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$=\frac{\left(\mathrm{2}{s}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$=\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }={L}.{H}.{S}. \\ $$

Commented by NECx last updated on 20/Oct/17

wow..... i′m most grateful.

$${wow}.....\:{i}'{m}\:{most}\:{grateful}. \\ $$

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