Question

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

• A. $\frac{2cR}{ac}$
• B. $\frac{2aR}{c^2}$
• C. $\frac{2aR}{bc}$
• D. None of these

Answer 1

The correct option is C $\frac{2aR}{bc}$

$b^2+c^2-a^2=2a^2$

$2bc\cos A=2a^2$

$\Rightarrow \cos A=\frac{2a^2}{2bc}=\frac{a^2}{bc}=\frac{{\sin }^{2}A}{\sin B\sin C}$

$\therefore \cot B+\cot C=\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C}$

$=\frac{\sin C\cos B+\cos C\sin B}{\sin B\sin C}$

$=\frac{\sin (B+C)}{\sin B\sin C}$

$=\frac{\sin (\pi -A)}{\sin B\sin C}$ since $A+B+C=\pi$

$=\frac{\sin A}{\sin B\sin C}$

Using sine rule

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

$\Rightarrow \sin A=\frac{a}{2R},\sin B=\frac{b}{2R},\sin C=\frac{c}{2R}$

$\frac{\sin A}{\sin B\sin C}$

$=\frac{\frac{a}{2R}}{\frac{b}{2R}\frac{c}{2R}}$

$=2R\frac{a}{bc}$