Questions: Find the exact value of the following expression, if possible. Do not use a calculator. tan^(-1)[tan(-π/6)] Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. tan^(-1)[tan(-π/6)]= (Simplify your answer. Type an exact answer, using π as needed. Use integers or fractions for any numbers in the expression.) B. There is no solution.

Solution Steps

To solve the expression \(\tan^{-1}[\tan(-\frac{\pi}{6})]\), we need to understand the properties of the tangent and arctangent functions. The tangent function is periodic with a period of \(\pi\), and the arctangent function returns values in the interval \((- \frac{\pi}{2}, \frac{\pi}{2})\). Since \(-\frac{\pi}{6}\) is within this interval, the arctangent of the tangent of \(-\frac{\pi}{6}\) will simply be \(-\frac{\pi}{6}\).

We need to find the exact value of the expression \(\tan^{-1}[\tan(-\frac{\pi}{6})]\). The tangent function is periodic with a period of \(\pi\), and the arctangent function returns values in the interval \((- \frac{\pi}{2}, \frac{\pi}{2})\).

First, we evaluate the inner function \(\tan(-\frac{\pi}{6})\). Since \(\tan(x)\) is periodic with period \(\pi\), we have: \[ \tan(-\frac{\pi}{6}) = -\tan(\frac{\pi}{6}) \] We know that: \[ \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} \] Thus: \[ \tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}} \]

Next, we apply the arctangent function to the result: \[ \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) \] Since \(-\frac{\pi}{6}\) is within the interval \((- \frac{\pi}{2}, \frac{\pi}{2})\), the arctangent of \(\tan(-\frac{\pi}{6})\) will simply be \(-\frac{\pi}{6}\).

Final Answer