In triangle ABC, the lengths of the three sides of the triangle are \displaystyle a\ cm,b\ cm\ and\ c\ cm . It is given that \displaystyle \frac{{{{a}^{2}}+{{b}^{2}}}}{{{{c}^{2}}}}=2016. Find the value of \displaystyle \frac{{\cot C}}{{\cot A+\cot B}} .
Solution
\displaystyle \frac{{{{a}^{2}}+{{b}^{2}}}}{{{{c}^{2}}}}=2016
\displaystyle {{a}^{2}}+{{b}^{2}}=2016\ {{c}^{2}}
\displaystyle {{a}^{2}}+{{b}^{2}}-{{c}^{2}}=2015\ {{c}^{2}}
\displaystyle 2ab\cos C=2015\ {{c}^{2}}\ \ (\because \ \ \cos C=\frac{{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}}{{2ab}})
\displaystyle \ \ \ \ \ \ \cos C=\frac{{2015\ {{c}^{2}}}}{{2ab}}
\displaystyle \frac{{\cot C}}{{\cot A+\cot B}}=\frac{{\frac{{\cos C}}{{\sin C}}}}{{\frac{{\cos A}}{{\sin A}}+\frac{{\cos B}}{{\sin B}}}}
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\frac{{\cos C}}{{\sin C}}}}{{\frac{{\sin B\cos A+\cos B\sin A}}{{\sin A\sin B}}}}
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\cos C}}{{\sin C}}\times \frac{{\sin A\sin B}}{{\sin (B+A)}}
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\cos C}}{{\sin C}}\times \frac{{\sin A\sin B}}{{\sin ({{{180}}^{\circ }}-C)}}
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\cos C}}{{\sin C}}\times \frac{{\sin A\sin B}}{{\sin C}}
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\cos C\times \frac{{\sin A}}{{\sin C}}\times \frac{{\sin B}}{{\sin C}}
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2015\ {{c}^{2}}}}{{2ab}}\times \frac{a}{c}\times \frac{b}{c}\ \ (\because \ \frac{a}{{\sin A}}=\frac{b}{{\sin B}}=\frac{c}{{\sin C}})
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2015}}{2}
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