1. First of all, a/sinA=b/sinB=c/sinC, so sinA+sinB=sinC(cosA+cosB) is equivalent to a+b=c(cosA+cosB).

And COSA = b 2+c 2-a 2/2bc, COSB = a 2+c 2-b 2/2ac,

So the original formula is equal to A+B = C (B2+C2-A2/2bc+A2+C2-B2/2ac).

The result is 2ab (a+b) = ab 2+AC 2+a 2b+BC 2-a 3-b 3.

After further sorting, AB 2+A 2b = AC 2+BC 2-A 3-B 3 is obtained.

So we get (a+b) (a 2+b 2-c 2) = 0.

Both A and B are positive numbers, and a+b must not be equal to 0, so A 2+B 2-C 2 = 0, so it is a right triangle, which may be better understood.

2. If the radius of inscribed circle is R, the area of △ABC is ab/2 or (ar+br+cr)/2. So ab/2=(ar+br+cr)/2,

Get ab = r (a+b+ 1), r = ab/a+b+ 1, when a=b, r reaches the maximum value of 1/2(√2+ 1), that is, √ 2-/kloc.

When a or b infinitely approaches 0, r infinitely approaches 0.

So 0

Hope to adopt