Cotangent Trigonometry

OK, We have to find the value of $\frac{\cot C}{\cot A + \cot B}$ Let us now simplify the numerator and denominator and bring them in terms of $\sin,\cos$ . $\mathscr{Numerator} ⇒ \frac{\cos C}{\sin C}$ $\mathscr{Denominator} ⇒ \frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}⇒\frac{\sin {C}}{\sin A \times \sin B}$

OK, Now we come to our simplified version of question that

$\frac{\cot C }{ \cot A + \cot B} = (\frac{\sin A}{\sin C } ) \times (\frac{\sin A}{\sin C} ) \times \cos C$

Now, by sine rule we know that $\frac{\sin A}{\sin C}=\frac{a}{c}$ , $\frac{\sin B}{\sin C}=\frac{b}{c}$ . And from cosine rule $\cos C=\frac{a^2+b^2-c^2}{2 \times a \times b}$ . Therefore Plugging all these results in our result, we get;;;

$(\frac{\sin A}{\sin C } ) \times (\frac{\sin A}{\sin C} ) \times \cos C =\frac{ab(a^2+b^2-c^2)}{2abc^2}=\frac{2011}{2}-\frac{1}{2}=1005$

SinA/a=SinB/b=SinC/c=k and CosA=(b^2+c^2-a^2)/2bc and so on then put in the question and get a easy product which is ((a^2+b^2)-c^2)/2c^2. then put the given values in equation.