Answer:
Solutions given:
4sin B=3sin (2A+B)............(1)
let B=A+B-A
and
2A+B=A+B+A
again
substituting value we get
Or, 4sin[ A+B-A]=3sin[ A+B+A]
Or,4Sin [(A+B)-A]=3sin[ (A+B)+A]
Or,4(Sin(A+B)cosA-Cos(A+B)sinA)=
3(Sin(A+B)cosA+Cos (A+B)sin A)
Or,4Sin(A+B)Cos A-4Cos(A+B)sinA=
3Sin(A+B)cosA+3cos(A+B)sinA
Or, 4Sin(A+B)CosA-3Sin(A+B)cosA=
3cos(A+B)sinA+4Cos(A+B)sinA
OrSin(A+B)CosA=7Cos(A+B)sinA
or
7cos(A+B)/Sin (A+B)=Cos A/Sin A
7Cot (A+B)=CotA.
7Cot (A+B)=CotA.Proved.
