Solving problem with Sines Law and Cosine Law

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}$