If sin⁡ α=A sin⁡ (α+β), A≠0, thenThe value of tan α isThe value of tan β isWhich of the following is not the value of tan⁡(α+β)?

sin⁡ α=A sin⁡ (α+β)=A(sin⁡ α cos⁡ β+sin⁡ β cos⁡ α)⇒ sin⁡ α (1−Acos⁡β)=A sin⁡ β cos⁡ α         …….. (i)⇒ tan⁡ α=A sin⁡ β(1−A cos⁡ β)                           ……… (ii)tan⁡β=sin⁡βcos⁡β=(1−Acos⁡β)tan⁡αAcos⁡β=(1−Acos⁡β)sin⁡αAcos⁡αcos⁡β             [from Eqs. (i) and (ii)]tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β =Asin⁡β1−Acos⁡β+sin⁡βcos⁡β1−Asin⁡βsin⁡β(1−Acos⁡β)cos⁡β =Asin⁡βcos⁡β+sin⁡β−Asin⁡βcos⁡βcos⁡β−Acos2⁡β−Asin2⁡β =sin⁡βcos⁡β−AAlso,   tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β =sin⁡αcos⁡α+sin⁡α(1−Acos⁡β)Acos⁡αcos⁡β1−sin2⁡α(1−Acos⁡β)Acos2⁡αcos⁡β                [frorn Eq. (ii)] =[Asin⁡αcos⁡β+sin⁡α−Asin⁡αcos⁡β]cos⁡αAcos2⁡αcos⁡β−sin2⁡α+Asin2⁡αcos⁡β =sin⁡αcos⁡αAcos⁡β−sin2⁡α