Question

In a \triangle ABC, if $b^2+c^2=3a^2,$ then cot B + cot C = cot A =

• A. 1
• B. $\frac{ab}{4△}$
• C. \N
• D. $\frac{ac}{4△}$

Answer 1

The correct option is C \N

$cotB+cotC-cotA=\frac{cosB}{sinB}+\frac{cosC}{sinC}-cotA$
$=\frac{sinCcosB+cosCsinB}{sinBsinC}-cosA=\frac{sin(B+C)}{sinBsinC}-\frac{cosA}{sinA}$
$=\frac{si{n}^{2}A-sinBsinCcosA}{sinAsinBsinC}=\frac{a^2-bccosA}{k\left(abc\right)}$
$Since=\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}=k\left(say\right)$
and cos $A=\frac{b^2+c^2-a^2}{2bc}=\frac{a^2-bc\frac{(b^2+c^2-a^2)}{2bc}}{\left(abc\right)k}$
$\frac{(a^2-a^2)}{abcK}=0,\{As\frac{b^2+c^2-a^2}{2}=\frac{3a^2-a^2}{2}=\frac{2a^2}{2}=a^2.\}$