If $$tan$$⁡θ$$tan$$⁡π$$3+$$θ$$tan$$⁡−π$$3+$$θ$$=ktan$$⁡$$3$$θ, then the value of k, is

We have, $$tan$$⁡θ$$tan$$⁡π$$3+$$θ$$tan$$⁡−π$$3+$$θ$$=ktan$$⁡$$3$$θ⇒ $$tan$$⁡θ$$tan$$⁡θ$$+31$$−$$3tan$$⁡θ$$tan$$⁡θ−$$31+3tan$$⁡θ$$=ktan$$⁡$$3$$θ⇒ $$tan3$$⁡θ−$$3tan$$⁡θ$$1$$−$$3tan2$$⁡θ$$=ktan$$⁡$$3$$θ⇒ −$$tan$$⁡$$3$$θ$$=ktan$$⁡$$3$$θ⇒$$k=$$−$$1$$

Questions

If $\tan θ \tan (\frac{π}{3} + θ) \tan (- \frac{π}{3} + θ) = k \tan 3 θ,$ then the value of $k,$ is

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-1

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detailed solution

Correct option is D

We have,

$\tan θ \tan (\frac{π}{3} + θ) \tan (- \frac{π}{3} + θ) = k \tan 3 θ$

$\begin{array}{l} \Rightarrow \tan θ (\frac{\tan θ + \sqrt{3}}{1 - \sqrt{3} \tan θ}) (\frac{\tan θ - \sqrt{3}}{1 + \sqrt{3} \tan θ}) = k \tan 3 θ \\ \Rightarrow \frac{\tan^{3} θ - 3 \tan θ}{1 - 3 \tan^{2} θ} = k \tan 3 θ \\ \Rightarrow - \tan 3 θ = k \tan 3 θ \Rightarrow k = - 1 \end{array}$