Question

In ΔABC, prove that (b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c.

Answer 1

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Please find below the solution to the asked query:

L.H.S.=(b−c)cos(A2)By sine rule we have,asinA=bsinB=csinC=k⇒L.H.S.=(ksinB−ksinC)cos(A2)=k(sinB−sinC)cos(A2)Using identity sinP−sinQ=2cos(P+Q2)sin(P−Q2), we get,L.H.S.=k{2cos(B+C2)sin(B−C2)}cos(A2)=k{2cos(B+C2)cos(A2)}sin(B−C2)Sum of all angles of triangle isπ.⇒A+B+C=π⇒B+C=π−A⇒B+C2=π2−A2⇒L.H.S.=k{2cos(π2−A2)cos(A2)}sin(B−C2)=k{2sin(A2)cos(A2)}sin(B−C2)Using identity 2sinα.cosα=sin2α, we get=ksinA.sin(B−C2)But ksinA=a⇒L.H.S.=asin(B−C2)=R.H.S. (Answer)

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