Question

- In a triangle ABC if tan A<0 then
A) tan B. tan C> 1
C) tan B. tan C = 1
B) tan B. tan C < 1
D) None
please explain​

Answer 1

Answer:

tanC<0

⇒

2

π

<C<π

⇒A+B<

2

π

tan(A+B+C)=

1−∑tanAtanB

∑tanA−ΠtanA

∵A+B+C=π

∑tanA=ΠtanA

tanAtanBtanC<0(∵

2

π

<C<π)

∴tanA+tanB+tanC<0

We have,

0<tan(A+B) ( As tanC<0 meansC is obtuse angle implies A+B acute)

Hence,

1−tanA×tanB

tanA+tanB

>0

⇒1−tanA×tanB>0

⇒tanA×tanB<1

please mark as brilliant