If α=π14 then the value of (tan⁡αtan⁡2α+tan⁡2αtan⁡4α+tan⁡4αtan⁡α) is

If $α = \frac{π}{14}$ then the value of $(\tan αtan 2 α + \tan 2 αtan 4 α + \tan 4 αtan α)$ is

A
1
B
$\frac{1}{2}$
C
2
D
$\frac{1}{3}$

Solution:

$α + 2 α + 4 α = 7 α = \frac{π}{2}$

$∴ \tan α \cdot \tan 2 α + \tan 2 α \cdot \tan 4 α + \tan α \cdot \tan 4 α = 1$

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