In a triangle ABC, if $\cos A=\cos B \cos C,$then $\tan B+\tan C=$

Solution:

$\cos B \cos C=\cos A=- \cos (B+C)$

$=-(\cos B \cos C-\sin B \sin C)$

$\Rightarrow 2\cos B \cos C=\sin B \sin C$

$\Rightarrow \tan B \tan C=2$

But $\tan A \tan B \tan C =\tan A+\tan B+\tan C$

$\therefore 2 \tan A=\tan A+\tan B+\tan C$

$\therefore \tan B+\tan C=\tan A$.