Solveeit Logo
Login

Question

Question: The value of \({{\tan }^{-1}}\left( \tan \dfrac{2\pi }{3} \right)\) is? (a) \(-\dfrac{\pi }{3}\) ...

The value of tan⁡−1(tan⁡2π3){{\tan }^{-1}}\left( \tan \dfrac{2\pi }{3} \right) is?
(a) −π3-\dfrac{\pi }{3}
(b) π3\dfrac{\pi }{3}
(c) −2π3-\dfrac{2\pi }{3}
(d) 2π3\dfrac{2\pi }{3}

Explanation

Solution

First of all write the argument of the tangent function which is 2π3\dfrac{2\pi }{3} as (π−π3)\left( \pi -\dfrac{\pi }{3} \right). Now, use the property of the tangent function given as tan⁡(π−θ)=−tan⁡θ\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(−x)=−tan⁡−1x{{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x to simplify. Finally, use the formula tan⁡−1(tan⁡x)=x{{\tan }^{-1}}\left( \tan x \right)=x for the domain values x∈(−π2,π2)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(tan⁡2π3){{\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,
⇒E=tan⁡−1(tan⁡2π3)\Rightarrow E={{\tan }^{-1}}\left( \tan \dfrac{2\pi }{3} \right)
Now, we know that the to use the formula tan⁡−1(tan⁡x)=x{{\tan }^{-1}}\left( \tan x \right)=x we must have the range of x given as x∈(−π2,π2)x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right). Clearly we can see that 2π3\dfrac{2\pi }{3} does not lie in the range (−π2,π2)\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 2π3\dfrac{2\pi }{3} as (π−π3)\left( \pi -\dfrac{\pi }{3} \right), so the expression becomes: -
⇒E=tan⁡−1(tan⁡(π−π3))\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⁡(π−θ)=−tan⁡θ\tan \left( \pi -\theta \right)=-\tan \theta we get,
⇒E=tan⁡−1(−tan⁡π3)\Rightarrow E={{\tan }^{-1}}\left( -\tan \dfrac{\pi }{3} \right)
Using the formula tan⁡−1(−x)=−tan⁡−1x{{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x we get,
⇒E=−tan⁡−1(tan⁡π3)\Rightarrow E=-{{\tan }^{-1}}\left( \tan \dfrac{\pi }{3} \right)
Now, we can see that π3\dfrac{\pi }{3} comes in the range (−π2,π2)\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right) so we can apply the formula tan⁡−1(tan⁡x)=x{{\tan }^{-1}}\left( \tan x \right)=x to get the answer, therefore we get,
∴E=−π3\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.