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Q.
In a $\triangle A B C$, if $b^{2}+c^{2}=3 a^{2}$, then $\cot B+\cot C-\cot A=$
Trigonometric Functions
Solution:
$\cot B+\cot C-\cot A=\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C}-\cot A$
$=\frac{\sin C \cos B+\cos C \sin B}{\sin B \sin C}-\cot A$
$=\frac{\sin (B+C)}{\sin B \sin C}-\frac{\cos A}{\sin A}$
$=\frac{\sin ^{2} A-\sin B \sin C \cos A}{\sin A \sin B \sin C}$
Since $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$ (say)
and $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$, we have
$\cot B+\cot C-\cot A=\frac{a^{2}-b c \frac{\left(b^{2}+c^{2}-a^{2}\right)}{2 b c}}{(a b c) k}$
$=\frac{\left(a^{2}-a^{2}\right)}{a b c k}=0,\left\{\right.$ As $\left.\frac{b^{2}+c^{2}-a^{2}}{2}=\frac{3 a^{2}-a^{2}}{2}=\frac{2 a^{2}}{2}=a^{2}\right\}$