Prove that
sin(A-B)/cosAcosB + sin(B-C)/cosBcosC + sin(C-A)/cosCcosA = 0
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
Ankit
To prove that
sin(A-B)/cosAcosB + sin(B-C)/cosBcosC + sin(C-A)/cosCcosA = 0
We can start by expanding the expression. Then we'll simplify it step by step to show that it equals zero.
1. Expand the expression:
sin(A-B)/cosAcosB + sin(B-C)/cosBcosC + sin(C-A)/cosCcosA
= [(sinAcosB - cosAsinB)/(cosAcosB)] + [(sinBcosC - cosBsinC)/(cosBcosC)] + [(sinCcosA - cosCsinA)/(cosCcosA)]
2. Simplify each term:
= [(sinAcosB - cosAsinB)/(cosAcosB)] + [(sinBcosC - cosBsinC)/(cosBcosC)] + [(sinCcosA - cosCsinA)/(cosCcosA)]
= [(sinAcosB)/(cosAcosB)] + [(-cosAsinB)/(cosAcosB)] + [(sinBcosC)/(cosBcosC)] + [(-cosBsinC)/(cosBcosC)] + [(sinCcosA)/(cosCcosA)] + [(-cosCsinA)/(cosCcosA)]
= [sinA/cosA] + [-sinB/cosB] + [sinB/cosB] + [-sinC/cosC] + [sinC/cosC] + [-sinA/cosA]
= sinA/cosA - sinA/cosA
= 0
Therefore, we have proved that
sin(A-B)/cosAcosB + sin(B-C)/cosBcosC + sin(C-A)/cosCcosA = 0.