Question

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

Answer 1

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.