If cosA+cosC=sinB then prove that ABC is right angled triangle

You mentioned that ABC is right angle triangle.Lets do some Hit and TrialB=90A=30C=60Then Cos A+ Cos C is not Equal to Sin BWhen we mention that ABC is a right angle triangle then B=90Hence we should mention that either ACB is a right angle triangle or CAB is right angle triangle. Then we can prove it

See as ABC is a triangle then, A+B+C=180° similarly, A+C=180°-B and B+C=180°-A Then, CosA+CosC=SinB2 Cos(A+C/2) * Cos(A-C/2)=2SinB/2 * CosB/22Cos(180°-B/2) * Cos(A-C/2)=2SinB/2 * CosB/2Cos(A-C/2)=CosB/2A-C=BA=B+CA=180°-A2A=180° Therefore, A=90°Hence proved

I am not doubting the answer, i also got it but what i mentioned is that if you say that ABC is a right angle triangle then by default B=90

See man if any angle in a triangle is 90° then the triangle is also right angled triangle. The triangle ABC can also be determined as: B / | / | / | /____ | C A

CosA+cosC=sinBCosA+cosC=cos(90-B)So A+C=B We know that A+B+C=180Now put A+C=BTherefore B+B=1802B=180 B=90