Question

In $\Delta ABC,$ if $b^2+c^2=2a^2,$ then the value of $\frac{\cot A}{\cot B+\cot C}$ is?

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

Answer 1

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

We know that in a $\Delta ABC$

$\cot A=\frac{R(b^2+c^2-a^2)}{abc}$
$\cot B=\frac{R(a^2+c^2-b^2)}{abc}$
$\cot C=\frac{R(b^2+a^2-c^2)}{abc}$
Where $R$ is a circumradius o the $\Delta ABC$
Therefore,
$\frac{\cot A}{\cot B+\cot C}$

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

Given:
$b^2+c^2=2a^2$

Substituting above, we get
$\frac{\cot A}{\cot b+\cot C}=\frac{2a^2-a^2}{2a^2}$
$=\frac{1}{2}$

Hence, option A.