`triangleABC` में
`a(sinB - sinC) + b(sinC - sinA) + c(sinA - sinB)= `

In a triangle ABC, prove that a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=0

In triangleABC, a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=

In any DeltaABC , prove that a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=0

If A+B+C=pi , prove that : (cosA)/(sinBsinC) + (cosB)/(sinC sinA) + (cosC)/(sinA sinB) =2 .

If in a triangle ABC , (sinA+sinB+sinC)(sinA+sinB-sinC)=3sinAsinB then

If A+B+C=pi , prove that : (cosA)/(sinb sinC) + (cosB)/(sinC sin) + (cosC)/(sinA sinB) =2 .

If A+B+C=pi , prove that : (cosA)/(sinb sinC) + (cosB)/(sinC sin) + (cosC)/(sinA sinB) =2 .

if A, B, C are acute positive angles, then : ((sinA + sinB)(sinA + sinC)(sinA + sinC))/(sinA sinB sinC) is :

Prove that (sin(B-C))/(sinB.sinC) + (sin(C-A))/(sinCsinA) + (sin(A-B))/(sinA.sinB) = 0

In any triangle ABC, prove that (a^(2)sin(B-C))/(sin B + sinC) + (b^(2)sin(C-A))/(sinC + sinA) +(c^(2)sin(A-B))/(sinA + sinB)=0