Question

In a acute triangle $\tan A\tan B+\tan B\tan C+\tan C\tan A>x$ then $x$ equals

• A. $3$
• B. $\sqrt{3}$
• C. $\frac{1}{3\sqrt{3}}$
• D. None of these

Answer 1

The correct option is D None of these

using the property in acute triangle,

$\sec A\sec B\sec C=\frac{1}{\cos A\cos B\cos C}$

above will be maximum at $A=B=C={60}^{\circ }$

$\therefore$ Max. $\left(\cos A\cos B\cos C\right)=\frac{1}{8}$

$\therefore$ Min. $\left(\sec A\sec B\sec C\right)=8$

Since we have $\cos (A+B+C)=\cos A\cos B\cos C(1-\tan A\tan B-\tan B\tan C-\tan C\tan A)$

Using the above calculated value of angle $A+B+C=\pi$ and $\cos \theta =-1,$

putting above values we get,

$\tan A\tan B\tan C+\tan C\tan A=1+\sec A\sec B\sec C$

$\therefore \tan A\tan B+\tan B\tan C+\tan C\tan A=9$

$\therefore \tan A\tan B+\tan B\tan C+\tan C\tan A>x$