Question

In a triangle ABC , if $b^2+c^2=3a^2$, then cot B + cot C -cot A is equal to

• A. $1$
• B. $\frac{ab}{4\Delta }$
• C. $0$
• D. $\frac{ac}{4\Delta }$

Answer 1

The correct option is C $0$

As we know

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

$=\frac{3a^2-(c^2+b^2)}{4\Delta }$

$=\frac{3a^2-3a^2}{4\Delta }=0$ (as $b^2+c^2=3a^2$)

Hence the answer is $0$.