Question

If $A+B+C=\pi$ and $\cos A=\cos B\cos C$, show that
$2\cot B\cot C=1$.

Answer 1

Given,

$\cos A=\cos B.\cos C$
or, $\sin A\cos A=\frac{\sin A}{\cos B\cos C}$
or, $\tan A=\frac{\sin (B+C)}{\cos B.\cos C}$
or, $\tan A=\frac{\sin B\cos C+\cos B\sin C}{\cos B\cos C}$
or, $\tan A=\tan B+\tan C$
Then $\tan (B+C)=\frac{\tan B+\tan C}{1-\tan B\tan C}$
or, $-\tan A=\frac{\tan A}{1-\tan B\tan C}$
or, $\tan B\tan C=2$
or, $\cot B\cot C=\frac{1}{2}$
or, $2\cot B.\cot C=1$.