If 3tan⁡θ−15∘=tan⁡θ+15∘, then θ is equal to

let A=θ+15∘,B=θ−15∘⇒A+B=2θ and A−B=30∘Now, tan⁡Atan⁡B=31or tan⁡A+tan⁡Btan⁡A−tan⁡B=3+13−1(applying componendo and dividendo rule) or  sin⁡(A+B)sin⁡(A−B)=2⇒sin⁡2θ=2sin⁡30∘=1⇒2θ=2nπ+π2 or θ=nπ+π4, n∈Z