If cos⁡(α+β)=4/5 and sin⁡(α−β)=5/13 where 0≤α,β≤π/4, then tan⁡2α=

0≤α,β≤π/4⇒ 0≤α+β≤π/2 ⇒−π/4≤α−β≤π/4Now cos⁡(α+β)=4/5 ⇒ tan⁡(α+β)=3/4and sin⁡(α−β)=5/13 ⇒ tan⁡(α−β)=5/12we have tan⁡2α=tan⁡[(α+β)+(α−β)]=tan⁡(α+β)+tan⁡(α−β)1−tan⁡(α+β)tan⁡(α−β)=(3/4)+(5/12)1−(3/4)(5/12)=14/1233/48=5633.