Question

If $A,B,C$ are the angles of a triangle such that the angle $A$ is obtuse, then $\tan B\tan C<$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

$\because A$ is obtuse angle.
Then $\frac{\pi }{2}<A<\pi$
$\Rightarrow \frac{\pi }{2}<\pi -(B+C)<\pi$
$\Rightarrow -\frac{\pi }{2}>-\pi +(B+C)>-\pi$
$\Rightarrow \frac{\pi }{2}>(B+C)>0$
$\because B+C<\frac{\pi }{2}$
$\Rightarrow B<\frac{\pi }{2}-C$
$\Rightarrow \tan B<\tan (\frac{\pi }{2}-C)$
$\Rightarrow \tan B<\cot C$
$\therefore \tan B<\frac{1}{\tan C}$
Hence $\tan B\tan C<1$