Question

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

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

Answer 1

Answer: (A)

$\tan A$

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

Since $A+B+C=\pi \Rightarrow A=\pi -(B+C)$
$\Rightarrow \cos B\cos C=\cos (\pi -(B+C))$
$\Rightarrow \cos B\cos C=-\cos (B+C)$
$\Rightarrow \cos B\cos C=-[\cos B\cos C-\sin B\sin C]$
$\Rightarrow 2\cos B\cos C=\sin B\sin C$

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

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

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