If A+B+C=180°, then (tanA+tanB+tanC)(tanAtanBtanC) =
0
2
1
-1
Explanation for correct option.
Given, A+B+C=180°
Now,
A+B+C=180°A+B=180°–CTaking‘tan’onbothsides,tan(A+B)=tan(180°–C)(tanA+tanB)(1–tanAtanB)=-tanCtanA+tanB=-tanC(1–tanAtanB)tanA+tanB=-tanC+tanAtanBtanCtanA+tanB+tanC=tanAtanBtanCtanA+tanB+tanCtanAtanBtanC=1
Hence, correct option is (C)