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`        In a triangle ABC, prove that acosA+bcosB+ccosC=2atanB×cosC`
2 years ago

```							Dear Aman Yku have typed a wrong RHSIt should be 2a sinB sinC. I have proved here-A/sinA=b/sinB=c/sinC=k (say)∴, a=ksinA, b=ksinB, c=ksinC∴, acosA+bcosB+ccosC=ksinAcosA+ksinBcosB+csinCcosC=k/2(2sinAcosA+2sinBcosB+2sinCcosC)=k/2(sin2A+sin2B+sin2C)=k/2[{2sin(2A+2B)/2cos(2A-2B)/2}+sin2C][∵, sinC+sinD=2sin(C+D)/2cos(C-D)/2]=k/2[2sin(A+B)cos(A-B)+2sinCcosC]=k[sin(π-C)cos(A-B)+sinCcos{π-(A+B)}]   [∵, A+B+C=π]=k[sinCcos(A-B)+sinC{-cos(A+B)}]=ksinC[cos(A-B)-cos(A+B)]   =ksinC[2sin(A-B+A+B)/2sin(A+B-A+B)/2][∵, cosC-cosD=2sin(C+D)/2sin(D-C)/2]=ksinC(2sinAsinB)=2(ksinA)sinBsinC=2asinBsinC  RegardsArun (askIITians forum expert)
```
2 years ago
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