If 3tan(θ−15∘)=tan(θ+15°), then θ=
nπ+π4
nπ+π8
nπ+π3
none of these
We have tan(θ+15)tan(θ−15)=31,
⇒sinθ+15cosθ-15sinθ-15cosθ+15=31
using componendo and dividendo
⇒sinθ+15cosθ-15+sinθ-15cosθ+15sinθ+15cosθ-15-sinθ-15cosθ+15=3+13-1
⇒sin2θsin300=2 ⇒sin2θ=1=sinπ2
⇒2θ=2nπ+π2 ⇒θ=nπ+π4