किसी `triangleABC` में सिद्ध करें कि
`a sinA - b sinB = c sin (A-B)`

Prove that a sinA - b sinB = c sin(A-B)

If sinA/sinB=sin(A-C)/sin(C-B) then

sin(A-B)=sinA cos B-cosA sinB

In triangleABC, (sinB)/(sin(A+B)) =

In triangleABC, (sinB)/(sin(A+B)) =

Statement:1 In triangleABC , if a lt b sinA , then the triangle is possible. And Statement:2 In triangleABC a/(sinA)= b/(sinB)

If sinA+cosB=a and sinB+cosA=b then sin(A+B)=

In a triangleABC , If tan A: tan B : tan C = 3:4:5, then the value of sinA.sinB.sinC is equal to:

यदि triangleABC में (If इन triangleABC), a = (1)/(sqrt(6)-sqrt(2)), b = (1)/(sqrt(6)+sqrt(2)), C = 60^(@) तो सिद्ध कीजिए कि (Prove that) c = (sqrt(4))/(2) .