Question

In triangle $ABC,a^2+c^2=2002b^2$, then $\frac{cotA+cotC}{cotB}$ is equal to

• A. $\frac{1}{2001}$
• B. $\frac{2}{2001}$
• C. $\frac{3}{2001}$
• D. $\frac{4}{2001}$

Answer 1

The correct option is B

$\frac{2}{2001}$

$\frac{cotA+cotC}{cotB}=\frac{\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C}}{\frac{\cos B}{\sin B}}=\frac{\sin (A+C)\sin B}{\sin A\cos B\sin C}=\frac{{\sin }^{2}B}{\sin A\cos B\sin C}=\frac{4R^2{b}^2}{4R^{2}ac\cos B}=\frac{2b^2}{2ac\cos B}=\frac{2b^2}{a^2+c^2-b^2}=\frac{2b^2}{2002b^2-b^2}=\frac{2}{2001}$