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:

When simplifying inverse trigonometric expressions, reduce each term using standard values and combine fractions carefully.

Answer 1

Step 1: Simplify each term. - First Term: \[ \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\tan^{-1}\left(\frac{1}{\sqrt{3}}\right). \] Since \( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \), the first term becomes: \[ -\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}. \] - Second Term: \[ \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) = \tan^{-1}\left(\sqrt{3}\right). \] Since \( \tan^{-1}\left(\sqrt{3}\right) = \frac{\pi}{3} \), the second term is: \[ \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3}. \] - 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: Add the simplified terms.
Combine the three terms: \[ -\frac{\pi}{6} + \frac{\pi}{3} - \frac{\pi}{4}. \] Find a common denominator (\( 12 \)) and simplify: \[ -\frac{\pi}{6} = -\frac{2\pi}{12}, \quad \frac{\pi}{3} = \frac{4\pi}{12}, \quad -\frac{\pi}{4} = -\frac{3\pi}{12}. \] Thus: \[ -\frac{\pi}{6} + \frac{\pi}{3} - \frac{\pi}{4} = -\frac{2\pi}{12} + \frac{4\pi}{12} - \frac{3\pi}{12} = -\frac{\pi}{12}. \] Step 3: Conclusion.
The value of the expression is: \[ \boxed{-\frac{\pi}{12}}. \]