Since α and β lies between θ and 4π and sin(α−β)>0, therefore, both α+β and α−β are positive acute angles.
Now, sin(α−β)=135 ⇒cos(α−β)=1−sin2(α−β) =1−(135)2=1312
and cos(α+β)=54 ⇒sin(α+β)=1−cos2(α+β) =1−(54)2=53.
Hence, tan(α+β)=cos(α+β)sin(α+β)=43
and tan(α−β)=cos(α−β)sin(α−β)=125.
Now, tan(2α)=tan{(α+β)+(α−β)}