If tanx2=tany3=tanz5,x+y+z=π and tan2x+tan2y+tan2z=K, then 3K ________ .
tanx=2t,tany=3t,tanz=5tAlso x+y+z=π∴tanx+tany+tanz=tanx tany tanz⇒t2=13⇒tan2x+tan2y+tan2z=t2(4+9+25)=38t2=383∴3K=38