In a triangle abc , if tan a less than 0 then prove that tan b.tan c less than 1

In a triangle ABC  we have from angle sum property  in Geometry  A+B+C =180 degree => A= 180-(B+C)  => tan A= tan(180-(B+C))= -tan(B+C) tan(B+C) >0 => (tan B +tan C)/(1- tanBTanC)>0

=> TanB+TanC>0 and 1-tanBtanC>0 ( This is the only possibility  for this rational trigonometric expression)

Now from second inequality it is clear that tanBxtanC