A maths

1 in triangle ABC, if cosA-cosB= sinC, prove that ABC is a right-angled triangle.
Since
cos A - cos B = sin C
-2sin[(A+B)/2]sin[(A-B)/2]=sin C
-2sin[(180-C)/2]sin[(A-B)/2]=sin C
-2cos(C/2)sin[(A-B)/2]=sin C
-2cos(C/2)sin[(A-B)/2]=2sin(C/2)[cos(C/2)]
-sin[(A-B)/2]=sin(C/2)
sin[(B-A)/2]=sin(C/2)
(B-A)/2=C/2
B-A=C
Because A + B + C = 180, A + B + B - A = 180, B=90
Prove that ABC is a right-angled triangle﹐B is a right-angled
2 if A+B+C=180^。 and 2sinB=sinA+sinC, prove that tan(A/2)tan(C/2)=1/3
2 sin B = sin A + sin C
2 sin(180-(A+C) )=2sin[(A+C)/2]cos[(A-C)/2]
sin(A+C)=sin[(A+C)/2]cos[(A-C)/2]
2sin[(A+C)/2]cos[(A+C)/2]=sin[(A+C)/2]cos[(A-C)/2]
2cos[(A+C)/2]=cos[(A-C)/2]
SO
tan (A/2) tan (C/2)
=[sin(A/2)sin(C/2)]/[cos(A/2)cos(C/2)]
={cos[(A-C)/2]-cos[(A+C)/2]}/{cos[(A-C)/2]+cos[(A+C)/2]}
={2cos[(A+C)/2]-cos[(A+C)/2]}/{2cos[(A+C)/2]+cos[(A+C)/2]}
=1/3