In `DeltaABC,(a+b+c)(tanA/2+tanB/2)=`

In DeltaABC, tanA=2, tanB=(3)/(2) and c=sqrt(65) , then circumradius of the triangle is : (a) 65 (b) (65)/(7) (c) (65)/(14) (d) none of these

In DeltaABC, tanA=2, tanB=(3)/(2) and c=sqrt(65) , then circumradius of the triangle is :

In DeltaABC , prove that: tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2=1

In DeltaABC , prove that: tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2=1

In a DeltaABC, if tanA+2tanB=0, then which one is correct?

In a DeltaABC, if tanA+2tanB=0, then which one is correct?

In DeltaABC , prove that: tanA/2.tanB/2.tanC/2=sqrt((1-a/s)(1-b/s)(1-c/s))

In a DeltaABC , if tanA/2=5/6 and tanB/2=20/37 then tanC/2=