In any triangle, prove that - a cosA + b cosB + c cosC​ = r/R
​​                                                         a + b + c

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Please find below the solution to the asked query:

L.H.S.=acosA+bcosB+ccosCa+b+cUse sine ruleasinA=bsinB=csinC=2RL.H.S.=2RsinA.cosA+2RsinB.cosB+2RsinC.cosC2RsinA+2RsinB+2RsinC=2sinA.cosA+2sinB.cosB+2sinC.cosC2sinA+sinB+sinC=sin2A+sin2B+sin2C2sinA+sinB+sinCUsing conditional identities of trigonometry we havesin2A+sin2B+sin2C=4sinA.sinB.sinCandsinA+sinB+sinC=4cosA2.cosB2.cosC2L.H.S.=4sinA.sinB.sinC24cosA2.cosB2.cosC2=sinA.sinB.sinC2cosA2.cosB2.cosC2=2sinA2.cosA2.2sinB2.cosB2.2sinC2.cosC22cosA2.cosB2.cosC2=4sinA2.sinB2.sinC2We know thatr=4RsinA2.sinB2.sinC24sinA2.sinB2.sinC2=rRHenceL.H.S.=rR=R.H.S. Hence Proved

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