Solution

LHS:

\[ 2\tan^{-1}(-3) = -2\tan^{-1}(3) \]

\[ = -\cos^{-1}\left(\frac{1 – 3^2}{1 + 3^2}\right) = -\cos^{-1}\left(\frac{-8}{10}\right) \]

\[ = -\cos^{-1}\left(\frac{-4}{5}\right) \]

\[ = -\left[\pi – \cos^{-1}\left(\frac{4}{5}\right)\right] \]

\[ = -\pi + \cos^{-1}\left(\frac{4}{5}\right) \]

Let \[ \cos^{-1}\left(\frac{4}{5}\right) = \theta \Rightarrow \cos\theta = \frac{4}{5} \Rightarrow \tan\theta = \frac{3}{4} \Rightarrow \theta = \tan^{-1}\left(\frac{3}{4}\right) \]

\[ = -\pi + \tan^{-1}\left(\frac{3}{4}\right) \]

\[ = -\pi + \left[\frac{\pi}{2} – \cot^{-1}\left(\frac{3}{4}\right)\right] \]

\[ = -\frac{\pi}{2} – \cot^{-1}\left(\frac{3}{4}\right) = -\frac{\pi}{2} – \tan^{-1}\left(\frac{4}{3}\right) \]

\[ = -\frac{\pi}{2} + \tan^{-1}\left(\frac{-4}{3}\right) \]

\[ \text{LHS} = \text{RHS} \quad \text{Hence proved.} \]