Question

If $\tan x=3\cot x$ find the value of x in radian.

Answer 1

Hint: Put cot x as $\dfrac{1}{\tan x}$ using the equation $\cot \theta =\dfrac{1}{\tan \theta }$ . Now, proceed further to get the value of tan x and hence, use the result:-
Value of $\tan {{60}^{\circ }}$ is $\sqrt{3}$ and value of $\tan {{120}^{\circ }}$ is $-\sqrt{3{}^\circ }$ use the following equation to get the angle (x) in radian form as:
$\pi \text{ radian}=180{}^\circ $ or
$1{}^\circ =\dfrac{\pi }{180{}^\circ }$ radian

Complete step-by-step answer:
The equation in the problem is $\tan x=3\cot x$ ………………………………. (1)
Hence, we need to find the value of x in radian from the above equation.
Now, as we know the relation between $\tan \theta $ and $\cot \theta $ is given as:
$\cot \theta =\dfrac{1}{\tan \theta }$ …………………………. (2)
So, we can replace cot x from the equation (1) with the help of above equation and can rewrite the equation (1) as:
$\tan x=3\times \dfrac{1}{\tan x}$ or
$\tan x=\dfrac{3}{\tan x}$
On cross-multiplying the above equation, we get ${{\tan }^{2}}x=3$
On taking square root to both the sides of the above equation, we get:
$\tan x=\pm \sqrt{3}$ ………………………… (3)
Case 1: $\tan x=\sqrt{3}$
Now, as we know value $\sqrt{3}$ can be given by tan function at angle ${{60}^{\circ }}$ . Hence, we get value of x in degree as:
$x={{60}^{\circ }}$ ………………………………. (4)
Case 2: $\tan x=-\sqrt{3}$
As we know the relation:
 $\tan \left( {{180}^{\circ }}-\theta \right)=-\tan \theta $ ……………………… (5)
And we also know that $\tan {{60}^{\circ }}=\sqrt{3}$ .
So, on putting $\theta ={{60}^{\circ }}$ to the equation (5), we get:
$\tan \left( {{180}^{\circ }}-{{60}^{\circ }} \right)=-\tan {{60}^{\circ }}$
$\tan {{120}^{\circ }}=-\sqrt{3}$………………….. (6)
Hence, we can write the relation $\tan x=-\sqrt{3}$ using above result as:
$\tan x=\tan {{120}^{\circ }}$
Or $x={{120}^{\circ }}$ ………………………. (7)
So, we get values of x as:
$x={{60}^{\circ }}$ and ${{120}^{\circ }}$
Now, as we know the relation between degree radian as:
$\pi \text{ radian }={{180}^{\circ }}$ or
${{180}^{\circ }}=\pi \text{ radian }$
So, we get:
$1{}^\circ =\dfrac{\pi }{180{}^\circ }\text{radian}$ ………………………..(8)
So, $x={{60}^{\circ }}$ in radian form is given as:
$x=\dfrac{\pi }{180}\times 60=\dfrac{\pi }{3}$ or
$x=\dfrac{\pi }{3}$
Similarly, $x=120{}^\circ $ in radian form is given as:
$\begin{align}
  & x=\dfrac{\pi }{180}\times 120=\dfrac{2\pi }{3} \\
 & x=\dfrac{2\pi }{3} \\
\end{align}$
Hence, values of x in radian form are given as:
$x=\dfrac{\pi }{3},\dfrac{2\pi }{3}$

Note: Another approach for the given problem would be that we can write tan x as $\dfrac{\sin x}{\cos x}$ and cot x as $\dfrac{\cos x}{\sin x}$ . so, we get –
$\dfrac{\sin x}{\cos x}=3\dfrac{\cos x}{\sin x}$
${{\sin }^{2}}x-3{{\cos }^{2}}x=0$
Now, use ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ to solve the problem further.
We know $\tan x=\sqrt{3}$ and $\tan x=-\sqrt{3}$ have infinite solutions and they can be given by the relation.
If, $\tan x=\tan y$ then $x=n\pi +y$ .
Where, $n\in z$
So, all the solutions of $\tan x=\sqrt{3}$ is given as:
$x=n\pi +\dfrac{\pi }{3},n\in z$ and for $\tan x=\sqrt{3}$ , we have:
$x-n\pi +\dfrac{2\pi }{3},n\in z$
So, we can respect x in the form of a general solution as well.