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$If \tan θ = \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \dots \infty}}}, where θ \in (0, 2 π), then the number of possible values of θ is$

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Correct option is C

Let tan⁡θ=x >0 since RHS of given equation is positive Then x=12+12+12+…..∞⇒x=12+x⇒x2+2x−1=0⇒x=−2±82=2−1 since x>0∴tan⁡θ=2−1⇒θ=π8 or 9π8

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If tan⁡θ=12+12+12+…∞ , where θ∈(0,2π) , then the number of possible values of θ is

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$If \tan θ = \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \dots \infty}}}, where θ \in (0, 2 π), then the number of possible values of θ is$