In a triangle tanA+tanB+tanC=6and tanA tanB=2.
Then the values of tanA,tanB and tanC are
1, 2, 3
2, 2, 2
1, 2, 0
None of these
Since A+B+C=π in triangle ABC, we have
⇒tanA+tanB+tanC=tanA tanB tanC⇒ 6=2tanC⇒ tanC=3 ∴ tanA+tanB=3, tanA tanB=2⇒ tanA=1 (or) 2 and tanB=2 (or) 1