Q.
Consider the following statements Statement I If 3tan(θ−15∘)=tan(θ+15∘) 0∘<θ<90∘, then θ=4π Statement II tan3θ=1−3tan2θ3tanθ−tan3θ
Choose the correct option.
Given that, 3tan(θ−15∘)=tan(θ+15∘) which can be rewritten as tan(θ−15∘)tan(θ+15∘)=13
Applying componendo and dividendo, we get tan(θ+15∘)−tan(θ−15∘)tan(θ+15∘)+tan(θ−15∘)=2 sin(θ+15∘)cos(θ−15∘)−sin(θ−15∘)cos(θ−15∘)sin(θ+15∘)cos(θ−15∘)+sin(θ−15∘)cos(θ+15∘)=2 ⇒sin30∘sin2θ=2 i.e., sin2θ=1 ⇒2θ=2π ∴θ=4π