Question

In a triangle $ABC,$ if $a^2+b^2=2011c^2,$ then $\frac{\cot C}{\cot A+\cot B}$ is equal to

• A. $999$
• B. $1002$
• C. $1005$
• D. $1006$

Answer 1

Answer: (C)

$1005$

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

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

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

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

$=\frac{\cos C}{\sin C}\times \frac{\sin A\sin B}{\sin (\pi -C)}$ since $A+B+C=\pi$

$=\frac{\cos C}{\sin C}\times \frac{\sin A\sin B}{\sin C}$

$=\frac{\cos C\sin A\sin B}{{\sin }^{2}C}$

$=\frac{ab\cos C}{c^2}$ by sine rule

$=\frac{1}{2c^2}(a^2+b^2-c^2)$

$=\frac{2011c^2-c^2}{2c^2}$

$=\frac{2010c^2}{2c^2}$

$=1005$