Solving problem with Sines law and Cosines law

In the diagram below , the lengths of the three sides of the triangle are $\displaystyle a\ cm\ ,\ b\ cm\ \text{and}\ c\ cm$ . It is given that $\displaystyle \frac{{{{a}^{2}}+{{b}^{2}}}}{{{{c}^{2}}}}=2011$ . Find the value of $\displaystyle \frac{{\cot C}}{{\cot A\ +\ \cot B}}$ .

Solution

$\displaystyle \frac{{{{a}^{2}}+{{b}^{2}}}}{{{{c}^{2}}}}=2011$

$\displaystyle {{a}^{2}}+{{b}^{2}}=2011\ {{c}^{2}}$

$\displaystyle {{a}^{2}}+{{b}^{2}}-{{c}^{2}}=2010\ {{c}^{2}}$

$\displaystyle \frac{{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}}{{2ab}}=\frac{{2010\ {{c}^{2}}}}{{2ab}}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \cos C=\frac{{2010\ {{c}^{2}}}}{{2ab}}\ \ \ \ \ \ \ \ \ \ \ \ \ \left[ {\because \cos C=\frac{{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}}{{2ab}}} \right]$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \cos C=1005\left( {\frac{c}{a}} \right)\left( {\frac{c}{b}} \right)$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \cos C=1005\left( {\frac{{\sin C}}{{\sin A}}} \right)\left( {\frac{{\sin C}}{{\sin B}}} \right)\ \ \ \left[ {\because \ \frac{a}{{\sin A}}=\frac{b}{{\sin B}}=\frac{c}{{\sin C}}} \right]$

$\displaystyle \ \ \ \ \ \ \ \ \ \frac{{\cos C}}{{\sin C}}=\frac{{1005\sin C}}{{\sin A\ \sin B}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \cot C=\frac{{1005\sin \left( {{{{180}}^{\circ }}-\left( {A+B} \right)} \right)}}{{\sin A\ \sin B}}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \cot C=\frac{{1005\sin \left( {A+B} \right)}}{{\sin A\ \sin B}}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \cot C=1005\left( {\frac{{\sin A\ \cos B+\cos A\ \sin B}}{{\sin A\ \sin B}}} \right)$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \cot C=1005\left( {\frac{{\sin A\ \cos B}}{{\sin A\ \sin B}}+\frac{{\cos A\ \sin B}}{{\sin A\ \sin B}}} \right)$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \cot C=1005\left( {\cot B+\cot A} \right)$ 

$\displaystyle \therefore \ \frac{{\cot C}}{{\cot A\ +\ \cot B}}=1005$