Question

cosA+2cosC/cosA+2cosB=sinB/sinC

Answer 1

your question is incomplete .
The complete question is ⇒ If (cosA+2cosC)/(cosA+2cosB)=sinB/sinc, prove that the triangle is either isosceles or right angled .

Proof : Given (cosA + 2cosC )/(cosA + 2cosB) = sinB/sinC
use componendo and dividendo rule ,
(2cosC - 2cosB)/(2cosA + 2cosB + 2cosC) = (sinB - sinC)/(sinB + sinC)
⇒(cosC - cosB)/(cosA + cosB + cosC) = (sinB - sinC)/(sinB + sinC)

Here it is clear that LHS = RHS , when (cosC - cosB) = 0 and (sinB - sinC) = 0
[ Note :- actually , I just take a possibility . Btw here many possibilities for solving it ]
so, (cosC - cosB) = 0 ⇒ cosC = cosB ⇒ C = B
also (sinB - sinC) = 0 ⇒ sinB = sinC ⇒ B = C

It means Given condition will be satisfy when B = C , means two angles are same of triangle . It means given condition will be possible when triangle is an isosceles .

again, Let A = 90° { right angle } so , (B + C)/2 = 45°
then, (cosC - cosB)/(cosC + cosB) = 2sin(C+B)/2.sin(B-C)/2/2cos(C+B)/2.cos(B-C)/2
= tan(B+C)/2.tan(B-C)/2 = tan45°.tan(B-C)/2 = tan(B-C)/2
(sinB - sinC)/(sinB + sinC) = 2cos(B+C)/2.sin(B-C)/2/2sin(B+C)/2.cos(B-C)/2
= cot(B+C)/2.tan(B-C)/2 = cot45°.tan(B-C)/2 = tan(B-C)/2
LHS = RHS

Hence, given condition will be possible when A = 90° , means triangle is a right angle triangle .

Hence, in both cases it is clear that When (cosA+2cosC)/(cosA+2cosB)=sinB/sinC
Then, triangle ABC is either isosceles or right angled triangle.