Question

If $A,B,C$ are angles of $\Delta ABC$ and $tanAtanC=3,tanBtanC=6$, then

• A. $A=\frac{\pi }{4}$
• B. $tanAtanB=2$
• C. $\frac{tanA}{tanC}=\frac{3}{2}$
• D. $ta{n}^{3}A+ta{n}^{3}B=-ta{n}^{3}C+3tanAtanBtanC=0$

Answer 1

The correct option is B $tanAtanB=2$

$\tan A+\tan B+\tan C=\tan AtanB\tan C=3\tan B=6\tan A$
$\therefore 3\tan A+\tan C=6\tan A\tan C=3\tan A$
$\therefore \tan A+2\tan A+3\tan A=\tan A\cdot 2tanA\cdot 3\tan A$
i.e. $\tan A={\tan }^{3}A$
i.e. ${\tan }^{2}A=1$ i.e. $\tan A=1$
$\therefore A=\frac{\pi }{4}$
$\therefore \tan A\tan B=2{\tan }^{2}A=2$