Answer:
[tex]\qquad\qquad\boxed{ \sf{ \: \: \bf \:\dfrac{tan(A + B)}{sinA} = \dfrac{3}{2} \: }} \\ \\ [/tex]
Step-by-step explanation:
Given that,
[tex]\sf \:sinB = \dfrac{1}{5}sin(2A + B) \\ \\ [/tex]
can be rewritten as
[tex]\sf \:\dfrac{sin(2A + B)}{sinB} = 5 \\ \\ [/tex]
[tex]\sf \:\dfrac{sin(2A + B)}{sinB} = \dfrac{5}{1} \\ \\ [/tex]
On applying Componendo and Dividendo, we get
[tex]\sf \:\dfrac{sin(2A + B) + sinB}{sin(2A + B) - sinB} = \dfrac{5 + 1}{5 - 1} \\ \\ [/tex]
can be further rewritten as
[tex]\sf \:\dfrac{2sin\bigg( \dfrac{2A + B + B}{2} \bigg)cos\bigg( \dfrac{2A + B - B}{2} \bigg) }{2cos\bigg( \dfrac{2A + B + B}{2} \bigg)sin\bigg( \dfrac{2A + B - B}{2} \bigg)} = \dfrac{6}{4} \\ \\ [/tex]
[tex]\sf \:\dfrac{sin\bigg( \dfrac{2A + 2B}{2} \bigg)cos\bigg( \dfrac{2A}{2} \bigg) }{cos\bigg( \dfrac{2A + 2B}{2} \bigg)sin\bigg( \dfrac{2A }{2} \bigg)} = \dfrac{3}{2} \\ \\ [/tex]
[tex]\sf \:\dfrac{sin(A + B)cosA}{cos(A + B)sinA} = \dfrac{3}{2} \\ \\ [/tex]
[tex]\sf\implies \: \sf \:\dfrac{tan(A + B)}{sinA} = \dfrac{3}{2} \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Formulae Used
[tex]\sf \:sinx +siny = 2sin\bigg( \dfrac{x + y}{2} \bigg)cos\bigg( \dfrac{x - y }{2} \bigg) \\ \\ [/tex]
[tex]\sf \:sinx - siny = 2cos\bigg( \dfrac{x + y}{2} \bigg)sin\bigg( \dfrac{x - y }{2} \bigg) \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\sf \:cosx +cosy = 2cos\bigg( \dfrac{x + y}{2} \bigg)cos\bigg( \dfrac{x - y }{2} \bigg) \\ \\ [/tex]
[tex]\sf \:cosx - cosy \: = \: - \: 2sin\bigg( \dfrac{x + y}{2} \bigg)sin\bigg( \dfrac{x - y }{2} \bigg) \\ \\ [/tex]