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If $3 \tan (θ - 15^{\circ}) = \tan (θ + 15^{°})$ , then $θ =$

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Correct option is A

We have tan(θ+15)tan(θ−15)=31,⇒sinθ+15cosθ-15sinθ-15cosθ+15=31using 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

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If 3tan(θ−15∘)=tan(θ+15°), then θ=

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If $3 \tan (θ - 15^{\circ}) = \tan (θ + 15^{°})$ , then $θ =$