First slide

Trigonometric equations

Question

The solution of $\tan x + \tan 2 x + \tan 3 x = \tan xtan 2 xtan 3 x$ is

Moderate

A

$x = nπ, n \in I$
B

$x = \frac{nπ}{2}, n \in I$
C

$x = \frac{nπ}{3}, n \in I$
D

None of these

Solution

we know that, $\tan 3 x = \tan (x + 2 x) = \frac{\tan x + \tan 2 x}{1 - \tan xtan 2 x}$
$\begin{array}{lr} \Rightarrow \tan 3 x - \tan xtan 2 xtan 3 x = \tan x + \tan 2 x \\ \Rightarrow \tan xtan 2 xtan 3 x = \tan 3 x - \tan x - \tan 2 x \dots (i) \\ Now, \tan x + \tan 2 x + \tan 3 x = \tan xtan 2 xtan 3 x \\ \Rightarrow \tan x + \tan 2 x + \tan 3 x = \tan 3 x - \tan x - \tan 2 x \\ \Rightarrow [from Eq. (i)] \\ \Rightarrow 2 x = nπ + (- x) \Rightarrow 3 x = nπ \\ \Rightarrow x = \frac{nπ}{3}, where n \in I \end{array}$

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