#(CosA+2CosC)/(CosA+2CosB)= SinB/SinC#, Prove that the triangle is either isosceles or right angled?

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Answer 1 · 2018-03-04T03:30:59

Given #rarr(cosA+2cosC)/(cosA+2cosB)=sinB/sinC#

#rarrcosAsinB+2sinB*cosB=cosAsinC+2sinCcosC#

#rarrcosAsinB+sin2B=cosAsinC+sin2C#

#rarrcosA(sinB-sinC)+sin2B-sin2C=0#

#rarrcosA[2sin((B-C)/2)*cos((B+C)/2)]+2*sin((2B-2C)/2)*cos((2B+2C)/2)]=0#

#rarrcosA[2sin((B-C)/2)*cos((B+C)/2)]+2*sin(B-C)*cos(B+C)]=0#

#rarrcosA[2sin((B-C)/2)*cos((B+C)/2)]+cosA*2*2*sin((B-C)/2)*cos((B-C)/2)]=0#

#rarr2cosA*sin((B-C)/2)[cos((B+C)/2)+2cos((B-C)/2)]=0#

Either, #cosA=0# #rarrA=90^@#

or, #sin((B-C)/2)=0# #rarrB=C#

Hence, the triangle is either isosceles or right angled. Credit goes to dk_ch sir.