If sin(B+C−A),sin(C+A−B),sin(A+B−C) are in A.P., then cot A, cotB, cotC are in
HP
Given:
sin(B+C−A),sin(C+A−B),sin(A+B−C)areinA.P⇒sin(C+A−B)−sin(B+C−A)=sin(A+B−C)−sin(C+A−B)⇒2sin(C+A−B−B−C+A2)COS(C+A−B+B+C−A2)=2sin(A+B−C−C−A+B2)cos(A+B−C+C+A−B2)⇒sin(A−B)cosC=sin(B−C)cosA⇒sinA cosB cosC−cosA sinB cosC⇒sinB cosC cosA−cosB sinC cosA⇒2sinB cosA cosC=sinA cosB cosC+cosA cosB sinCDividingbothsidesbycosA cosB cosC:2tanB=tanA+tanC⇒2cotB=1cotA+1cotC