Sin( B-C)/ cosBcosC + sin(C-A)/ cosCcosA + sin( A-B)/cosAcosB = 0 Pl solve the problem faster Mark u brainleist
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Answer:
Step-by-step explanation:
Sin( B-C)/ cosBcosC + sin(C-A)/ cosCcosA + sin( A-B)/cosAcosB
//We know that Sin(A-B) = SinACosB - SinBCosA
=> SinBCosc - SinCCosB/CosBCosC + SinCCosA - SinACosC/CosCCosA
+ SinACosB - SInBCosA/CosACosB
=> TanB - TanC + TanC - TanA + TanA - TanB
=> 0
= R.H.S
Hence proved.
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