
The value of ${{\tan }^{-1}}\left( \tan \dfrac{2\pi }{3} \right)$ is?
(a) $-\dfrac{\pi }{3}$
(b) $\dfrac{\pi }{3}$
(c) $-\dfrac{2\pi }{3}$
(d) $\dfrac{2\pi }{3}$
Answer
458.4k+ views
Hint: First of all write the argument of the tangent function which is $\dfrac{2\pi }{3}$ as $\left( \pi -\dfrac{\pi }{3} \right)$. Now, use the property of the tangent function given as $\tan \left( \pi -\theta \right)=-\tan \theta $ by considering the fact that the tangent function is negative in the second quadrant. Further, use the property of the inverse tangent function given as ${{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x$ to simplify. Finally, use the formula ${{\tan }^{-1}}\left( \tan x \right)=x$ for the domain values $x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ to get the answer.
Complete step-by-step solution:
Here we have been provided with the expression ${{\tan }^{-1}}\left( \tan \dfrac{2\pi }{3} \right)$ and we are asked to find its value. let us assume the expression as E so we have,
$\Rightarrow E={{\tan }^{-1}}\left( \tan \dfrac{2\pi }{3} \right)$
Now, we know that the to use the formula ${{\tan }^{-1}}\left( \tan x \right)=x$ we must have the range of x given as $x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$. Clearly we can see that $\dfrac{2\pi }{3}$ does not lie in the range $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ so we cannot use the given formula now but first we need to simplify the expression. We know that we can write $\dfrac{2\pi }{3}$ as $\left( \pi -\dfrac{\pi }{3} \right)$, so the expression becomes: -
$\Rightarrow E={{\tan }^{-1}}\left( \tan \left( \pi -\dfrac{\pi }{3} \right) \right)$
We know that the tangent function is negative in the second quadrant so using the formula $\tan \left( \pi -\theta \right)=-\tan \theta $ we get,
$\Rightarrow E={{\tan }^{-1}}\left( -\tan \dfrac{\pi }{3} \right)$
Using the formula ${{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x$ we get,
$\Rightarrow E=-{{\tan }^{-1}}\left( \tan \dfrac{\pi }{3} \right)$
Now, we can see that $\dfrac{\pi }{3}$ comes in the range $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ so we can apply the formula ${{\tan }^{-1}}\left( \tan x \right)=x$ to get the answer, therefore we get,
$\therefore E=-\dfrac{\pi }{3}$
Hence, option (a) is the correct answer.
Note: Note that you cannot directly remove the inverse trigonometric and trigonometric functions if the angle doesn’t lie in the defined range. You may get the answer wrong while doing such. Remember the domain values of all the inverse trigonometric functions and the formulas related to them. You must also remember the signs of all the six trigonometric functions in all the four quadrants.
Complete step-by-step solution:
Here we have been provided with the expression ${{\tan }^{-1}}\left( \tan \dfrac{2\pi }{3} \right)$ and we are asked to find its value. let us assume the expression as E so we have,
$\Rightarrow E={{\tan }^{-1}}\left( \tan \dfrac{2\pi }{3} \right)$
Now, we know that the to use the formula ${{\tan }^{-1}}\left( \tan x \right)=x$ we must have the range of x given as $x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$. Clearly we can see that $\dfrac{2\pi }{3}$ does not lie in the range $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ so we cannot use the given formula now but first we need to simplify the expression. We know that we can write $\dfrac{2\pi }{3}$ as $\left( \pi -\dfrac{\pi }{3} \right)$, so the expression becomes: -
$\Rightarrow E={{\tan }^{-1}}\left( \tan \left( \pi -\dfrac{\pi }{3} \right) \right)$
We know that the tangent function is negative in the second quadrant so using the formula $\tan \left( \pi -\theta \right)=-\tan \theta $ we get,
$\Rightarrow E={{\tan }^{-1}}\left( -\tan \dfrac{\pi }{3} \right)$
Using the formula ${{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x$ we get,
$\Rightarrow E=-{{\tan }^{-1}}\left( \tan \dfrac{\pi }{3} \right)$
Now, we can see that $\dfrac{\pi }{3}$ comes in the range $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ so we can apply the formula ${{\tan }^{-1}}\left( \tan x \right)=x$ to get the answer, therefore we get,
$\therefore E=-\dfrac{\pi }{3}$
Hence, option (a) is the correct answer.
Note: Note that you cannot directly remove the inverse trigonometric and trigonometric functions if the angle doesn’t lie in the defined range. You may get the answer wrong while doing such. Remember the domain values of all the inverse trigonometric functions and the formulas related to them. You must also remember the signs of all the six trigonometric functions in all the four quadrants.
Recently Updated Pages
Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Why is the cell called the structural and functional class 12 biology CBSE

Sketch the electric field lines in case of an electric class 12 physics CBSE

State and explain Coulombs law in electrostatics class 12 physics CBSE

a Tabulate the differences in the characteristics of class 12 chemistry CBSE
