Find the exact value of \( \tan \left[2 \cos ^{-1}\left(-\frac{3}{5}\right)\right] \) \( \tan \left[2 \cos ^{-1}\left(-\frac{3}{5}\right)\right]=\square \) (Simplify your answer. Type an exact answer, using radicals as needed. Use integers or fractions for any numbers in the expression.)

To find \( \tan \left[2 \cos ^{-1}\left(-\frac{3}{5}\right)\right] \), we start by letting \( \theta = \cos^{-1}\left(-\frac{3}{5}\right) \). This means \( \cos(\theta) = -\frac{3}{5} \). We can use the identity for \( \tan(2\theta) \): \[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \] Next, we need to find \( \tan(\theta) \). From the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), we have: \[ \sin^2(\theta) + \left(-\frac{3}{5}\right)^2 = 1 \] \[ \sin^2(\theta) + \frac{9}{25} = 1 \] \[ \sin^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25} \] So, \( \sin(\theta) = \frac{4}{5} \) (we use the positive root since \( \theta \) corresponds to the range where sine is positive). Now we can find \( \tan(\theta) \): \[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3} \] Next, substituting this into the double angle formula for tangent: \[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} = \frac{2\left(-\frac{4}{3}\right)}{1 - \left(-\frac{4}{3}\right)^2} \] Calculating \( \tan^2(\theta) \): \[ \tan^2(\theta) = \left(-\frac{4}{3}\right)^2 = \frac{16}{9} \] Now substituting this back: \[ \tan(2\theta) = \frac{-\frac{8}{3}}{1 - \frac{16}{9}} = \frac{-\frac{8}{3}}{\frac{9}{9} - \frac{16}{9}} = \frac{-\frac{8}{3}}{-\frac{7}{9}} = \frac{-8}{3} \cdot \frac{-9}{7} \] Simplifying this gives: \[ \tan(2\theta) = \frac{72}{21} = \frac{24}{7} \] Therefore, the exact value is: \[ \tan \left[2 \cos^{-1}\left(-\frac{3}{5}\right)\right] = \frac{24}{7} \]