If tan⁡θtan⁡π3+θtan⁡−π3+θ=ktan⁡3θ, then the value of k, is

A
1
B
$1 / 3$
C
$3$
D
-1

Solution:

We have,

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

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

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

Solution:

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