If tan⁡(α−β)=sin⁡2β3−cos⁡2β , then

we have sin⁡2β3−cos⁡2β=2sin⁡β⋅cos⁡β2−2cos⁡2β+1+cos⁡2β=2sin⁡β⋅cos⁡β4sin2⁡β+2cos2⁡β=tan⁡β1+2tan2⁡β=2tan⁡β−tan⁡β1+2tan2⁡β=tan⁡(α−β)∴ tan⁡α−tan⁡β1+tan⁡α⋅tan⁡β=2tan⁡β−tan⁡β1+2tan2⁡β∴ tan⁡α=2tan⁡β