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

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

Grade:12th pass

1 Answers

Arun
25763 Points
4 years ago
Dear Aman
 
Yku have typed a wrong RHS
It 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 
 
Regards
Arun (askIITians forum expert)

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