Question

In a $△ABC,$ sides $a,b,c$ are in $A.P$ and $\frac{2}{1!9!}+\frac{2}{3!7!}+\frac{1}{5!5!}=\frac{8^a}{\left(2b\right)!}$, then the maximum value of $\tan A\tan B$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

Answer: (B)

$\frac{1}{3}$

$\frac{2}{1!9!}+\frac{2}{3!7!}+\frac{1}{5!5!}=\frac{8^a}{\left(2b\right)!}$
$\to \frac{1}{1!9!}+\frac{1}{3!7!}+\frac{1}{5!5!}+\frac{1}{1!9!}+\frac{1}{3!7!}=\frac{8^a}{\left(2b\right)!}$
$\Rightarrow \frac{1}{10!}(\frac{10!}{1!9!}+\frac{10!}{3!7!}+\frac{10!}{5!5!}+\frac{10!}{1!9!})=\frac{8^a}{\left(2b\right)!}$
$\Rightarrow \frac{1}{10!}\left(2^9\right)=\frac{8^a}{\left(2b\right)!}$
$\Rightarrow \frac{1}{10!}\left(2^9\right)=\frac{2^{3a}}{\left(2b\right)!}$
Equating the like terms, we get
$\Rightarrow 3a=9$ and $2b=10$
$\therefore a=3$ and $b=5$
Also, $2b=a+c\Rightarrow 10=3+c\Rightarrow c=7$
$\therefore a=3,b=5$ and $c=7$
$\because \frac{\tan A+\tan B}{2}\ge \sqrt{\tan A\tan B}$ .............$\left(1\right)$
Also,$\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{9+25-49}{30}=\frac{-1}{2}$ which is in second quadrant.
$\therefore C=180-60=120$degrees.
and $\angle A,\angle B<{60}^0$ (given)
$\tan A+\tan B+\tan C=\tan A\tan B\tan C$
$\Rightarrow \tan A+\tan B-\sqrt{3}=\tan A\tan B(-\sqrt{3})$
$\Rightarrow \tan A+\tan B=\sqrt{3}(1-\tan A\tan B)$ ...............$\left(2\right)$
Also, $\tan A+\tan B>0$
$\Rightarrow \sqrt{3}(1-\tan A\tan B)>0$
$\Rightarrow 1-\tan A\tan B>0$
$\Rightarrow \tan A\tan B<1$ ...............$\left(3\right)$
From $\left(1\right)$ and $\left(2\right)$ we have
$\frac{\sqrt{3}(1-\tan A\tan B)}{2}\ge \sqrt{\tan A\tan B}$
Let $\tan A\tan B=\lambda$
$\therefore \sqrt{3}(1-\lambda )\ge 2\sqrt{\lambda }$
Squaring both sides we get
$\Rightarrow 3{(1-\lambda )}^2\ge 4\lambda$
$\Rightarrow 3{\lambda }^2-10\lambda +3\ge 0$
$\Rightarrow (3\lambda -1)(\lambda -3)\ge 0$
$\because \lambda -3<0$
$\therefore 3\lambda -1\le 0$
$\Rightarrow \lambda \le \frac{1}{3}$
Hence, $\tan A\tan B\le \frac{1}{3}$