Question

value of tan inverse (-1/root3)

Answer 1

-π/6

Answer 2

To find the value of $\tan^{-1}(-\frac{1}{\sqrt{3}})$, we need to determine the principal value of the inverse tangent function.

The principal value branch for $\tan^{-1}x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Let $y = \tan^{-1}(-\frac{1}{\sqrt{3}})$. This implies that $\tan y = -\frac{1}{\sqrt{3}}$.

We know that $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$. Since $\tan y$ is negative, and the principal value of $\tan^{-1}x$ lies in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, the angle $y$ must be in the fourth quadrant. In the fourth quadrant, we use the identity $\tan(-\theta) = -\tan(\theta)$. Therefore, $\tan(-\frac{\pi}{6}) = -\tan(\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}$.

Since $-\frac{\pi}{6}$ lies within the principal value branch $(-\frac{\pi}{2}, \frac{\pi}{2})$, the value of $\tan^{-1}(-\frac{1}{\sqrt{3}})$ is $-\frac{\pi}{6}$.