Evaluating Inverse Trigonometric Expression
Find the value of:
[NCERT, Ex. 2.3, Q.5, Page.35]
\[
\tan^{-1}\left( \tan \frac{2\pi}{3} \right)
\]
Solution
\[
\tan^{-1}\left( \tan \frac{2\pi}{3} \right)
\]
We begin with the given expression.
\[
= \tan^{-1} \left[ \tan \left( \pi – \frac{\pi}{3} \right) \right]
\]
Express \( \frac{2\pi}{3} \) as \( \pi – \frac{\pi}{3} \).
\[
= \tan^{-1} \left( -\tan \frac{\pi}{3} \right)
\]
Using the identity \( \tan(\pi – \theta) = -\tan \theta \).
\[
= -\tan^{-1} \left( \tan \frac{\pi}{3} \right)
\]
Since \( \tan^{-1}(-x) = -\tan^{-1}(x) \).
\[
= -\frac{\pi}{3}
\]
Evaluate the inverse trigonometric function.
Important Note:
Remember that \( \tan^{-1}\left( \tan \frac{2\pi}{3} \right) \ne \frac{2\pi}{3} \).
This is because \( \tan^{-1}(\tan x) = x \) only if \( x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \),
and \( \frac{2\pi}{3} \notin \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).
\[
\therefore \tan^{-1}\left( \tan \frac{2\pi}{3} \right) = -\frac{\pi}{3}
\]