$tan x + tan 2 x + tan 3 x = 0$ then x =

n π + \frac{π}{3}

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n π + \frac{π}{4}

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n π

n π \pm \frac{π}{3}

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2 n π

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Solution

The correct option is B $n π$ or $n π \pm \frac{π}{3}$
$L H S = \frac{sin x}{cos x} + \frac{sin 2 x}{cos 2 x} + \frac{sin 3 x}{cos 3 x}$
LCM of cosx and cos2x is taken,so
$= \frac{sin x cos 2 x + cos x sin 2 x}{cos x cos 2 x} + \frac{sin 3 x}{cos 3 x}$

$= \frac{sin 3 x}{cos x cos 2 x} + \frac{sin 3 x}{cos 3 x}$

$sin 3 x = 0$ or $cos 3 x + cos x cos 2 x = 0$
$sin 3 x = 0$ or $2 cos 3 x + 2 cos x cos 2 x = 0$
$sin 3 x = 0$ or $2 cos 3 x + cos 3 x + cos x = 0$
$sin 3 x = 0$ or $3 cos 3 x + cos x = 0$
$sin 3 x = 0$ or $3 (4 {cos}^{3} x - 3 cos x) + cos x = 0$
$x = n π$ or $x = n π \pm \frac{π}{3}$

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tanx+tan2x+tan3x=0 then x =

$tan x + tan 2 x + tan 3 x = 0$ then x =