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Q. In a triangle $A B C$ if $\tan A<0$ then:

Trigonometric Functions

Solution:

$\because \quad \tan A<0$ and $A+B+C=180^{\circ}$
$\Rightarrow A >90^{\circ} $
$\Rightarrow B+C< 90^{\circ}$
$\Rightarrow \tan (B+C) >0 $
$ \Rightarrow \frac{\tan B+\tan C}{1-\tan B \tan C}>0 $
$\Rightarrow 1-\tan B \tan C >0 $
$ \Rightarrow \tan B \tan C<1 $