Asked by Kewal
Prove that
sin(A-B)/cosAcosB + sin(B-C)/cosBcosC + sin(C-A)/cosCcosA = 0
sin(A-B)/cosAcosB + sin(B-C)/cosBcosC + sin(C-A)/cosCcosA = 0
Answers
Answered by
Reiny
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
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
Answered by
Anonymous
Ankit
Answered by
manish yadav allahabad
good
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