# Solve the Following Equation: Tan X + Tan 2 X = Tan 3 X - Mathematics

Sum

Solve the following equation:

$\tan x + \tan 2x = \tan 3x$

#### Solution

Given:

$\tan x + \tan2x = \tan3x$
Now,
$\tan x + \tan2x = \tan (x + 2x)$
$\Rightarrow \tan x + \tan 2x = \frac{\tan x + \tan2x}{1 - \tan x \tan2x}$
$\Rightarrow \tan x + \tan2x - \frac{\tan x + \tan2x}{1 - \tan x \tan2x} = 0$
$\Rightarrow (\tan x + \tan2x) (1 - \tan x \tan2x) - (\tan x + \tan2x) = 0$
$\Rightarrow (\tan x + \tan 2x) (1 - \tan x \tan2x - 1) = 0$
$\Rightarrow (\tan x + \tan2x) ( - \tan x \tan2x) = 0$
$\Rightarrow \tan x + \tan 2x = 0$ or
$\tan x \tan2x = 0$

Now,

$\tan x + \tan 2x = 0$
$\Rightarrow \tan x = - \tan 2x$
$\Rightarrow \tan x = \tan - 2x$
$\Rightarrow x = n\pi - 2x, n \in Z$
$\Rightarrow 3x = n\pi$
$\Rightarrow x = \frac{n\pi}{3}, n \in Z$

And,

$\tan x + \tan 2x = 0$
$\Rightarrow \tan x = - \tan 2x$
$\Rightarrow \tan x = \tan - 2x$
$\Rightarrow x = n\pi - 2x, n \in Z$
$\Rightarrow 3x = n\pi$
$\Rightarrow x = \frac{n\pi}{3}, n \in Z$

∴ $x = \frac{n\pi}{3}, n \in Z$ or

$x = \frac{n\pi}{3}, n \in Z$
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 11 Trigonometric equations
Exercise 11.1 | Q 5.2 | Page 22