Question

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

• A. ${\tan }^{2}A$
• B. $\tan A$
• C. $\tan B$
• D. $\cos A$

Answer 1

The correct option is B $\tan A$

Given:$\cos B\cos C=\cos A$

$\Rightarrow \cos B\cos C=\cos (\pi -(B+C))$

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

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

$\Rightarrow \cos B\cos C=-\cos B\cos C+\sin B\sin C$

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

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

$\Rightarrow 2=\frac{\sin B\sin C}{\cos B\cos C}$

$\Rightarrow \tan A\tan B=2$

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

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

$\Rightarrow \tan A-\tan A=\tan B+\tan C$

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

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