Question

Help meeee!!!!...!!!!!!!!!​

Answer 1

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.