Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $a^2+c^2=2002b^2,$ then $\frac{\cot A+\cot C}{\cot B}$ 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 $\frac{2}{2001}$

In a triangle $ABC$

$cotA=\frac{b^2+c^2-a^2}{4\Delta }$

$cotB=\frac{c^2+a^2-b^2}{4\Delta }$

and $cotC=\frac{a^2+b^2-c^2}{4\Delta }$

From the above formulas,
$\frac{cotA+cotC}{cotB}$
$=\frac{(a^2+b^2-c^2)+(b^2+c^2-a^2)}{c^2+a^2-b^2)}$
$=\frac{2b^2}{2002b^2-b^2}$
$=\frac{2}{2001}$.