Question

In $\Delta ABC$, if $b^2+c^2=2a^2$, then value of $\frac{cotA}{cotB+cotC}$ is

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $\frac{5}{2}$
• D. $\frac{5}{3}$

Answer 1

The correct option is A $\frac{1}{2}$

$\cot A=\frac{R(b^2+c^2-a^2)}{abc}$
$\cot B=\frac{R(a^2+c^2-b^2)}{abc}$
$\cot C=\frac{R(b^2+a^2-c^2)}{abc}$
Hence
$\frac{cotA}{\cot B+\cot C}$
$=\frac{b^2+c^2-a^2}{a^2+b^2-c^2+a^2+c^2-b^2}$
$=\frac{b^2+c^2-a^2}{2a^2}$
Now
$b^2+c^2=2a^2$
Substituting above, we get
$=\frac{2a^2-a^2}{2a^2}$
$=\frac{1}{2}$