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)
