Question

In $△ABC,a^2+c^2=2002b^2$, then $\frac{\cot A+\cot C}{\cot B}$ equals to

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

Answer 1

Answer: (B)

$\frac{2}{2001}$

$a^2+c^2=2002b^2$

By sine Rule $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

$\frac{a}{2R}=\sin A$

$\frac{b}{2R}=\sin B$

$\frac{c}{2R}=\sin C$

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

$=\frac{\frac{b^2+c^2-a^2}{abc}+\frac{a62+b^2-c^2}{abc}}{\frac{a^2+c^2-b^2}{abc}}$

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

$=\frac{2b^2}{2002b^2-b^2}$

$=\frac{2b^2}{2001b^2}$

$\frac{\cot A+\cot C}{\cot B}=\frac{2}{2001}$