In a triangle ABC we define
x=tanB−C2tanA2,y=tanC−A2tanB2 and z=tanAB2tanC2
Then the value of x+y+z (in terms of x,y,z) is
-xyz
tanB−C2=b−cb+ccotA2, Napier's analogy∴x1=tanB−C2tanA2=b−cb+c⇒xb+xc=b−c⇒c(x+1)b(1−x)∴1+x1−x=bcSimilarly 1+y1−y=ca and 1+z1−z=ab∴(1+x)(1+y)(1+z)(1−x)(1−y)(1−z)=bc.ca.ab=1∴(1+x)(1+y+z+yz)(1−x)(1−y−z+yz)=1⇒1+y+z+yz+x+xy+xz+xyz=1−y−z+yz−x+xy+xz−xyz⇒2(x+y+z)=−2xyz∴(x+y+z)=−xyz