Question

In a triangle ABC, tan C < 0. Then

• (A) tan A . tan B < 1
• (B) tan A . tan B > 1
• (C) tan A + tan B + tan C < 0
• (D) tan A + tan B + tan C > 0

Answer 1

The Correct Option is A, D

Explanation:
Here, $\tan C<0\Rightarrow C$ is obtuse.$\Rightarrow A$ and $B$ are acute and $A+B<\frac{\pi }{2}$ $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}>0$$\Rightarrow 1-\tan A\cdot \tan B>0\Rightarrow \tan A\tan B<1$$\tan A+\tan B+\tan C=(+)(+)(-)<0=\tan A\cdot \tan B\cdot \tan C\mathrm{in}\mathrm{\Delta }ABC$