**Answer:**

**sin(α+β)/sin(α-β) ==(tan α+tan β)/(tan α-tan β )
**

**Step-by-step explanation:**

We have to complete

**sin(α+β)/sin(α-β) = ?
**

The identities that will be used:

**sin(α+β)=sin α cos β+cos α sin β
**

**and
**

**sin(α-β)=sin α cos β-cos α sin β **

Now:

= sin(α+β)/sin(α-β)

=(sin α cos β+cos α sin β)/(sin α cos β-cos α sin β)

In order to bring the equation in compact form we wil divide both numerator and denominator with cos α cos β

= (((sin α cos β+cos α sin β))/(cos α cos β ))/(((sin α cos β-cos α sin β))/(cos α cos β))

=((sin α cosβ)/(cos α cos β )+(cos α sin β)/(cos α cos β ))/((sin α cos β)/(cos α cos β )-(cos α sin β)/(cos α cos β))

=(sin α/cos α + sin β/cos β )/(sin α/cos β - sin β/cos β)

=(tan α+tan β)/(tan α-tan β )

So,

**sin(α+β)/sin(α-β) ==(tan α+tan β)/(tan α-tan β)**

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