If $tan x + tan 2 x + tan 8 x = tan x . tan 2 x . tan 8 x$ then the general solution of $x =$

n π, \forall n ϵ Z

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n π, \frac{π}{4}, \forall n ϵ Z

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\frac{n π}{11}, \forall n ϵ Z

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n π, \frac{π}{3}, \forall n ϵ Z

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Solution

The correct option is C $\frac{n π}{11}, \forall n ϵ Z$
Given

$tan x + tan 2 x + tan 8 x = tan x tan 2 x tan 8 x$

Consider

$tan (x + 2 x + 8 x) = \frac{tan x + tan 2 x + tan 8 x - tan x tan 2 x tan 8 x}{1 - tan x tan 2 x - tan x tan 8 x - tan 2 x tan 8 x}$

$⟹ tan (x + 2 x + 8 x) = \frac{tan x tan 2 x tan 8 x - tan x tan 2 x tan 8 x}{1 - tan x tan 2 x - tan x tan 8 x - tan 2 x tan 8 x}$

$⟹ tan (x + 2 x + 8 x) = 0$

$⟹ tan 11 x = 0$

$⟹ 11 x = n π$

$⟹ x = \frac{n π}{11}$

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