Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively, and $a^2+c^2=2002b^2$, 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 $\frac{2}{2001}$

$\frac{\cot A+\cot C}{\cot B}=\frac{\frac{bc\cos A}{2\Delta }+\frac{ab\cos C}{2\Delta }}{\frac{ac\cos B}{2\Delta }}=\frac{b(c\cos A+a\cos C)}{ac\cos B}=\frac{2b\left(b\right)}{a^2+c^2-b^2}=\frac{2}{\frac{a^2+c^2}{b^2}-1}..........\left(i\right)$

Given $\frac{a^2+c^2}{b^2}=2002$

Substituting in $\left(i\right)$

$=\frac{2}{2002-1}=\frac{2}{2001}$