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

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\tan A

2 \tan A

c o t A

\tan^{2} A

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detailed solution

Correct option is A

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

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

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

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

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

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

$∴ \tan B + \tan C = \tan A$ .

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In a triangle ABC, if cosA=cosB cosC,then tanB+tanC=