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# If $\tan x+\tan 2x+\tan 3x=0$ , then $x=?$ (a) $n\pi +\dfrac{\pi }{3}$ (b) $n\pi +\dfrac{\pi }{4}$ (c) $\dfrac{n\pi }{3}$ or ${{\tan }^{-1}}\left( \pm \dfrac{1}{\sqrt{2}} \right)$ (d) $2n\pi$

Last updated date: 10th Aug 2024
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Hint: We will split the angle $3x$ as $x+2x$ . We will use the formula for the tan function of the sum of two angles. This formula is given by $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$ . Then we will substitute the expression for $\tan x+\tan 2x$ in the given equation. After that we will use the formula for the double angle, which is given by $\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$ and simplify the equation to obtain the plausible value of $x$ .

The given equation is $\tan x+\tan 2x+\tan 3x=0$ . Let us split the angle $3x$ as $x+2x$ . We know that the formula for the tangent of sum of angles is given by
$\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$
Substituting $a=x$ and $b=2x$ in the above formula, we get the following expression,
\begin{align} & \tan \left( x+2x \right)=\dfrac{\tan x+\tan 2x}{1-\tan x\tan 2x} \\ & \therefore \tan 3x\left( 1-\tan x\tan 2x \right)=\tan x+\tan 2x \\ \end{align}
Now, we will substitute the value of $\tan x+\tan 2x$ in the given equation. So, we get the following expression,
$\tan 3x\left( 1-\tan x\tan 2x \right)+\tan 3x=0$
Simplifying the above equation by taking $\tan 3x$ common, we get
\begin{align} & \tan 3x\left( 1-\tan x\tan 2x+1 \right)=0 \\ & \therefore \tan 3x\left( 2-\tan x\tan 2x \right)=0 \\ \end{align}
From the above equation, we can conclude that either $\tan 3x=0$ or $2-\tan x\tan 2x=0$ .
We know that $\tan n\pi =0$ . Therefore, if $\tan 3x=0$ , then we have $n\pi =3x$ . Hence, $x=\dfrac{n\pi }{3}$ .
Next, if $2-\tan x\tan 2x=0$ then, we will simplify the equation in the following manner,
$\tan x\tan 2x=2$
We know the double angle formula for the tan function. It is given by $\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$ . Substituting this value in place of the double angle in the above equation, we get
\begin{align} & \tan x\left( \dfrac{2\tan x}{1-{{\tan }^{2}}x} \right)=2 \\ & \Rightarrow \dfrac{2{{\tan }^{2}}x}{1-{{\tan }^{2}}x}=2 \\ & \Rightarrow {{\tan }^{2}}x=1-{{\tan }^{2}}x \\ & \Rightarrow 2{{\tan }^{2}}x=1 \\ & \Rightarrow {{\tan }^{2}}x=\dfrac{1}{2} \\ & \therefore \tan x=\pm \dfrac{1}{\sqrt{2}} \\ \end{align}
From the above equation, we have $x={{\tan }^{-1}}\left( \pm \dfrac{1}{\sqrt{2}} \right)$ .
Hence, we have $x=\dfrac{n\pi }{3}$ or $x={{\tan }^{-1}}\left( \pm \dfrac{1}{\sqrt{2}} \right)$ . Therefore, the correct option is C.

Note:
It is useful to know the formulae for double angles and the sum of angles for all trigonometric functions. These formulae help us to simplify equations. The calculations should be done explicitly so that we can avoid making any errors and obtain the correct answer. It is beneficial to have the knowledge of inverse trigonometric functions for such types of questions as well.