Question

sin(A-B)/cosA.cosB+ sin(B-C)/CosB.cosC+sin(C-A)/cosc.cosA=0​

Answer 1

Answer:

Sin(A-B)/cosA.cosB + sin (B-C)/cosB.cosC +sin (C-A)/cosC.cosA …

Homework Help

Sin(A-B)/cosA.cosB + sin (B-C)/cosB.cosC +sin (C-A)/cosC.cosA =0

Posted by Jishan Khan 3 years, 1 month ago

2 answers

Rashmi Bajpayee 3 years, 1 month ago

sin(A−B)cosAcosB+sin(B−C)cosBcosC+sin(C−A)cosCcosA

= sinAcosB−cosAsinBcosAcosB+sinBcosC−cosBsinCcosBcosC+sinCcosA−cosCsinAcosCcosA

= sinAcosBcosC−cosAsinBcosC+cosAsinBcosC−cosAcosBsinC+sinCcosAcosB−cosBcosCsinAcosAcosBcosC

= 0cosAcosBcosC

= 0

Hence proved

Step-by-step explanation:

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