Question

Find the value of \( \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) + \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) + \tan^{-1}\left[\sin\left(-\frac{\pi}{2}\right)\right]. \)

Hint:

To simplify inverse trigonometric expressions, convert each term to a common denominator when adding or subtracting fractions.

Answer 1

Step 1: Simplify each term individually. - First Term: \[ \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\tan^{-1}\left(\frac{1}{\sqrt{3}}\right). \] Since \( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \), we have: \[ -\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}. \] - Second Term: \[ \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) = \tan^{-1}(\sqrt{3}). \] We know that \( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \), so the second term becomes: \[ \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3}. \] - Third Term: \[ \tan^{-1}\left[\sin\left(-\frac{\pi}{2}\right)\right] = \tan^{-1}(-1). \] Since \( \tan^{-1}(-1) = -\frac{\pi}{4} \), the third term is: \[ \tan^{-1}(-1) = -\frac{\pi}{4}. \] Step 2: Combine the simplified terms.
Now, we add the three terms: \[ -\frac{\pi}{6} + \frac{\pi}{3} - \frac{\pi}{4}. \] To add these, we first convert each term to have a common denominator. The common denominator is \( 12 \): \[ -\frac{\pi}{6} = -\frac{2\pi}{12}, \quad \frac{\pi}{3} = \frac{4\pi}{12}, \quad -\frac{\pi}{4} = -\frac{3\pi}{12}. \] Thus, the sum is: \[ -\frac{2\pi}{12} + \frac{4\pi}{12} - \frac{3\pi}{12} = -\frac{\pi}{12}. \] Step 3: Conclusion.
The value of the expression is: \[ \boxed{-\frac{\pi}{12}}. \]