Question

In $△ABC$ if the joining orthocenter and centroid is parallel to $CA$, then $\tan A\tan A=$

• A. $2$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $3$

Answer 1

The correct option is A $2$

$\begin{array}{c}\text{In}△ABC,C\text{is the circumcentre and}0\text{is the}\\ \text{orthocentre. Extend}C\text{and}0\text{to meet}AC\text{at}P\text{and}Q\\ \text{We lunow that}\\ CP=R\cos B\\ OQ=2R\cos A\cos C\\ \Rightarrow CP=QQ\\ \Rightarrow R\cos B=2R\cos A\cos C\\ \Rightarrow \cos B=2\cos A\cos C\end{array}$

$\begin{array}{c}\Rightarrow \cos (\pi -(A+E\left)\right)=2\cos A\cos C\\ \Rightarrow 2\cos A\cos C=-\cos (A+C)\\ \Rightarrow -(\cos A\cos C+\sin A\sin C)\\ \Rightarrow 3\cos A\cos C=+\sin A\sin C\\ \Rightarrow \tan A\tan C=3\\ \text{hence,}\left(D\right)\text{is the correct option}\end{array}$