Question

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

If $b^2+c^2=2a^2,$ then value of $\frac{cotA}{\cot B+\cot C}$ is?

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

Answer 1

Answer: (A)

$\frac{1}{2}$

we can write

$\frac{cotA}{cotB+cotC}=\frac{\frac{b^2+c^2-a^2}{4△}}{\frac{c^2+a^2-b^2+a^2+b^2-c^2}{4△}}=\frac{b^2+c^2-a^2}{2a^2}$

since, $b^2+c^2=2a^2$

$\frac{b^2+c^2-a^2}{2a^2}=\frac{2a^2-a^2}{2a^2}=\frac{a^2}{2a^2}=\frac{1}{2}$

Therefore, Answer is $A$