Question

Prove that
sin(A-B)/cosAcosB + sin(B-C)/cosBcosC + sin(C-A)/cosCcosA = 0

Answer 1

for the LS, form a common denominator of
cosAcosBcosC.

LS = [cosC(sinAcosB - cosAsinB) + cosA(sinBcosC - cosBsinC) + cosB(sinCcosA - cosCsinA) ]/(cosAcosBcosC)

expand the top, it will add up to zero
= 0/(cosAcosBcosC)
= 0
= RS

Answer 2

Ankit

Answer 3

good