Trigonometry! #71

$b^{2}+c^{2}=3a^{2}$

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

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

$\cos A=\frac{a^{2}}{bc}$

Also $\sin A=\frac{a}{k}$

$\Rightarrow \cot A=\frac{\frac{a^{2}}{bc}}{\frac{a}{k}}=\frac{ak}{bc}$

Now, $\cot B+\cot C$

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

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

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

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

$=\frac{\frac{a}{k}}{\frac{b}{k}\cdot\frac{c}{k}}$

$=\frac{ak}{bc}$

$\therefore \cot B+\cot C=\cot A$

$\Rightarrow \frac{\cot B+\cot C}{\cot A}=1$