Given \( f(x)=-\tan ^{2}(x) \), write the equation of the line tangent to \( y=f(x) \) when \( x=\frac{4 \pi}{3} \)

To find the equation of the tangent line to \( y = f(x) = -\tan^2(x) \) at \( x = \frac{4\pi}{3} \), we first need to calculate \( f\left(\frac{4\pi}{3}\right) \) and then find the derivative \( f'(x) \). 1. Calculate \( f\left(\frac{4\pi}{3}\right) \): \[ f\left(\frac{4\pi}{3}\right) = -\tan^2\left(\frac{4\pi}{3}\right) = -\left(\tan\left(\frac{4\pi}{3}\right)\right)^2 \] Since \( \tan\left(\frac{4\pi}{3}\right) = \tan\left(\pi + \frac{\pi}{3}\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \), \[ f\left(\frac{4\pi}{3}\right) = -(\sqrt{3})^2 = -3. \] 2. Find \( f'(x) \): Using the chain rule, \[ f'(x) = -2\tan(x) \sec^2(x). \] Now, we need to evaluate \( f'\left(\frac{4\pi}{3}\right) \): \[ f'\left(\frac{4\pi}{3}\right) = -2\tan\left(\frac{4\pi}{3}\right) \sec^2\left(\frac{4\pi}{3}\right) = -2(-\sqrt{3})\sec^2\left(\frac{4\pi}{3}\right). \] Knowing \( \sec^2 = 1 + \tan^2 \), \[ \sec^2\left(\frac{4\pi}{3}\right) = 1 + (\sqrt{3})^2 = 4. \] Therefore, \[ f'\left(\frac{4\pi}{3}\right) = -2(-\sqrt{3})(4) = 8\sqrt{3}. \] 3. Now we have the point \(\left(\frac{4\pi}{3}, -3\right)\) and the slope \(8\sqrt{3}\). The equation of the tangent line in point-slope form is: \[ y - (-3) = 8\sqrt{3}\left(x - \frac{4\pi}{3}\right). \] Simplifying gives: \[ y + 3 = 8\sqrt{3}x - \frac{32\pi\sqrt{3}}{3}. \] Rearranging gives the tangent line's equation: \[ y = 8\sqrt{3}x - \frac{32\pi\sqrt{3}}{3} - 3. \] Thus, the equation of the tangent line is: \[ y = 8\sqrt{3}x - \frac{32\pi\sqrt{3}}{3} - 3. \]