Question:
Evaluate : \( \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) + \cot^{-1\left(\frac{1}{\sqrt{3}}\right) + \tan^{-1}\left(\sin\left(-\frac{\pi}{2}\right)\right) + \tan^{-1}\left(\tan\frac{2\pi}{3}\right) \)}
Evaluate : \( \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) + \cot^{-1\left(\frac{1}{\sqrt{3}}\right) + \tan^{-1}\left(\sin\left(-\frac{\pi}{2}\right)\right) + \tan^{-1}\left(\tan\frac{2\pi}{3}\right) \)}
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Always verify if the angle inside \( \tan^{-1}(\tan \theta) \) is within the principal range before simplifying.
Remember: \( \tan^{-1}(-x) = -\tan^{-1}x \).
Remember: \( \tan^{-1}(-x) = -\tan^{-1}x \).
Updated On: Sep 10, 2026
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Solution and Explanation
Concept:
• The principal value range of \( \tan^{-1}x \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).
• The principal value range of \( \cot^{-1}x \) is \( (0, \pi) \).
• For \( \tan^{-1}(\tan \theta) = \theta \), \( \theta \) must lie in \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). If it does not, we use trigonometric identities to find an equivalent angle within the range.
Step 1: Evaluate each inverse trigonometric term individually
Term 1: \( \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) \)
We know that \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \). Since the range is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \):
\( \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} \)
Term 2: \( \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) \)
We know that \( \cot \frac{\pi}{3} = \frac{1}{\sqrt{3}} \). Since the range is \( (0, \pi) \):
\( \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3} \)
Term 3: \( \tan^{-1}\left(\sin\left(-\frac{\pi}{2}\right)\right) \)
First, evaluate the inner term: \( \sin\left(-\frac{\pi}{2}\right) = -1 \).
Then, \( \tan^{-1}(-1) = -\frac{\pi}{4} \).
Term 4: \( \tan^{-1}\left(\tan\frac{2\pi}{3}\right) \)
The angle \( \frac{2\pi}{3} \) is outside the principal range \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).
Using \( \tan(\pi - \theta) = -\tan \theta \):
\( \tan\left(\frac{2\pi}{3}\right) = \tan\left(\pi - \frac{\pi}{3}\right) = -\tan\frac{\pi}{3} = -\sqrt{3} \).
So, \( \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \).
Step 2: Sum all the evaluated values
Substitute the values back into the original expression:
\[ \text{Expression} = \left(-\frac{\pi}{6}\right) + \left(\frac{\pi}{3}\right) + \left(-\frac{\pi}{4}\right) + \left(-\frac{\pi}{3}\right) \] Cancel the \( \frac{\pi}{3} \) and \( -\frac{\pi}{3} \) terms:
\[ \text{Expression} = -\frac{\pi}{6} - \frac{\pi}{4} \]
Step 3: Find a common denominator and simplify
The least common multiple of 6 and 4 is 12:
\[ \text{Expression} = \frac{-2\pi - 3\pi}{12} = -\frac{5\pi}{12} \]
• The principal value range of \( \tan^{-1}x \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).
• The principal value range of \( \cot^{-1}x \) is \( (0, \pi) \).
• For \( \tan^{-1}(\tan \theta) = \theta \), \( \theta \) must lie in \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). If it does not, we use trigonometric identities to find an equivalent angle within the range.
Step 1: Evaluate each inverse trigonometric term individually
Term 1: \( \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) \)
We know that \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \). Since the range is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \):
\( \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} \)
Term 2: \( \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) \)
We know that \( \cot \frac{\pi}{3} = \frac{1}{\sqrt{3}} \). Since the range is \( (0, \pi) \):
\( \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3} \)
Term 3: \( \tan^{-1}\left(\sin\left(-\frac{\pi}{2}\right)\right) \)
First, evaluate the inner term: \( \sin\left(-\frac{\pi}{2}\right) = -1 \).
Then, \( \tan^{-1}(-1) = -\frac{\pi}{4} \).
Term 4: \( \tan^{-1}\left(\tan\frac{2\pi}{3}\right) \)
The angle \( \frac{2\pi}{3} \) is outside the principal range \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).
Using \( \tan(\pi - \theta) = -\tan \theta \):
\( \tan\left(\frac{2\pi}{3}\right) = \tan\left(\pi - \frac{\pi}{3}\right) = -\tan\frac{\pi}{3} = -\sqrt{3} \).
So, \( \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \).
Step 2: Sum all the evaluated values
Substitute the values back into the original expression:
\[ \text{Expression} = \left(-\frac{\pi}{6}\right) + \left(\frac{\pi}{3}\right) + \left(-\frac{\pi}{4}\right) + \left(-\frac{\pi}{3}\right) \] Cancel the \( \frac{\pi}{3} \) and \( -\frac{\pi}{3} \) terms:
\[ \text{Expression} = -\frac{\pi}{6} - \frac{\pi}{4} \]
Step 3: Find a common denominator and simplify
The least common multiple of 6 and 4 is 12:
\[ \text{Expression} = \frac{-2\pi - 3\pi}{12} = -\frac{5\pi}{12} \]
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