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Q.

If $\tan θ . \tan (120 ° - θ) \tan (120 ° + θ) = \frac{1}{\sqrt{3}}$ , then $θ =$

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a

\frac{n π}{3} - \frac{π}{12}, n \in z

b

\frac{n π}{3} - \frac{π}{18}, n \in z

c

\frac{n π}{3} + \frac{π}{18}, n \in z

d

\frac{n π}{3} + \frac{π}{12}, n \in z

answer is C.

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Detailed Solution

\tan θ . \tan (120 ° - θ) \tan (120 ° + θ) = \frac{1}{\sqrt{3}}

$\tan 3 θ = \frac{1}{\sqrt{3}} = \tan \frac{π}{6}$

$3 θ = n π + \frac{π}{6}$

$θ = \frac{n π}{3} + \frac{π}{18}$

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If tanθ.tan(120°−θ)tan(120°+θ)=13 , then θ=