Question

The general solution of $\dfrac{{\tan 2x - \tan x}}{{1 + \tan x\tan 2x}} = 1$ is
A. $n\pi + \dfrac{\pi }{4},\forall n \in Z$
B. $n\pi \pm \dfrac{\pi }{4},\forall n \in Z$
C. $\phi $
D. $n\pi + \dfrac{\pi }{6},\forall n \in Z$

Answer 1

Hint – We will start solving this question by using some tangent formula. We will also take some help from trigonometric ratios tables to find some values and after solving them step by step, we will get our solution.

Complete step-by-step answer:

The given trigonometric equation is,
$\dfrac{{\tan 2x - \tan x}}{{1 + \tan x\tan 2x}} = 1$

This equation can also be written as,
$\dfrac{{\tan 2x - \tan x}}{{1 + \tan 2x\tan x}} = 1$…………. (1)

As we know that in equation (1) $\dfrac{{\tan 2x - \tan x}}{{1 + \tan 2x\tan x}}$ represents a formula , i.e.,
$\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$

Therefore, equation (1) becomes,
$\tan \left( {2x - x} \right) = 1$
$ \Rightarrow \tan x = 1$
We know that tangent is 1 at $\dfrac{\pi }{4}$, therefore the above equation becomes,
$\tan x = \tan \dfrac{\pi }{4}$
Also, we know that if $\theta $ and $\alpha $ are not the multiples of $\dfrac{\pi }{2}$, then
$
  \tan \theta = tan\alpha \\
   \Rightarrow \theta = n\pi + \alpha ,\forall n \in Z \\
 $
 where $Z$ represents integers.
Thus, $x = n\pi + \dfrac{\pi }{4},\forall n \in Z$, is the general solution of the given equation.

Hence, option A is the right answer.
Note – The expression involving integer n which gives all solutions of trigonometric equations is called a general solution. These questions are complex so they should be solved with caution and all the formulas and properties should be on tips.