`a(b cosC-c cosB)=` ................

a(cosC-cosB)=2(b-c)cos^2A/2

If A,B and C are angles of a triangle, then the determinant |{:(-1,cosC,cosB),(cosC,-1,cosA),(cosB,cosA,-1):}| is equal to "........."

(a+b+c)(cosA+cosB+cosC)

Prove that (sin(B-C))/(cosB cosC)+(sin(C-A))/(cosC cosA)+(sin(A-B))/(cosA cosB) =0 .

For any angles A,B,C. (sin(A-B))/(cosA.cosB)+(sin(B-C))/(cosB.cosC)+(sin(C-A))/(cosC.cosA)=...

In DeltaABC , prove that: (c-bcosA)/(b-c (cosA))=(cosB)/(cosC)

In any A B C ,2(bc cosA+ ca cosB+ab cosC) =

Prove that b cosB + c cosC = a cos ( B-C)

In DeltaABC , prove that: a(cosC-cosB)=2(b-c)cos^(2)A/2

In DeltaABC , prove that: a(cosC-cosB)=2(b-c)cos^(2)A/2