Question

Complete the identity

Answer 1

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

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 cos⁡β)/(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)