Question

if sin B= 3sin(2A+B) prove that 2tanA+ tan(A+B)=0​

Answer 1

Answer:

2tanA+ tan(A+B)=0​

Step-by-step explanation:

Given

sin B= 3sin(2A+B)

B = A + B - A  

& 2A + B = A + B + A

=> Sin(A + B - A  ) = 3 Sin(A + B + A)

Using

Sin(C + D) = SinCCosD + CosCSinD

& SIn(C - D) = SinCCosD - CosCSinD

=> Sin(A+B)CosA - Cos(A+B)SinA = 3 (Sin(A+B)CosA + Cos(A+B)CosA)

=> 2 Sin(A+B)CosA  + 4Cos(A+B)SinA =  0

Diving both side 2CosACos(A+ B)

=> 2 Sin(A+B)CosA/2CosACos(A+ B)  + 4Cos(A+B)SinA /2CosACos(A+ B) = 0

=> Tan(A + B) + 2TanA = 0

=> 2tanA+ tan(A+B)=0​

Answer 2

Answer:

Step-by-step explanation:

Refer to the attachment given below ❤️

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