If tan3θ−1tan3θ+1=√3, then the general value of θ is
A
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B
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C
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D
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Solution
The correct option is C Given that, tan3θ−1tan3θ+1=√3 ⇒tan3θ−1=√3tan3θ+√3 ⇒tan3θ−1−√3tan3θ−√3=0 ⇒tan3θ(1−√3)−(1+√3)=0 ⇒tan3θ=1+√31−√3 =tan45∘+tan60∘1−tan45∘.tan60∘ =tan(45∘+60∘) ⇒tan3θ=tan105∘=tan7π12 ∴3θ=nπ+7π12⇒θ=nπ3+7π36