Question

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Answer 1

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sinB=3sin(2A+B)

⇒sin(A+B−A)=3sin(A+A+B)

⇒sin(A+B)cosA−cos(A+B)sin(B)=3[sin(A+B)cosA+cos(A+B)sin(B)]

          divide both side by cos(A+B).cosA

⇒tan(A+B)−tan(A)=3tan(A+B)+3tan(A)

⇒4tan(A)+2tan(A+B)=0

∴2tan(A)+tan(A+B)=0

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