Prove that: \(2\tan^{-1}\left(\dfrac13\right) + \cos^{-1}\left(\dfrac35\right) = \dfrac{\pi}{2}\).

Answer: Proved.

Using \(2\tan^{-1}x = \tan^{-1}\dfrac{2x}{1 - x^2}\) for \(|x| \lt 1\):

\(2\tan^{-1}\dfrac13 = \tan^{-1}\dfrac{2/3}{1 - 1/9} = \tan^{-1}\dfrac{2/3}{8/9} = \tan^{-1}\dfrac34\).

Let \(\cos^{-1}\dfrac35 = \alpha\), \(0 \lt \alpha \lt \frac{\pi}{2}\). Then \(\cos\alpha = \frac35\), \(\sin\alpha = \frac45\), so \(\tan\alpha = \frac43\) and \(\cos^{-1}\dfrac35 = \tan^{-1}\dfrac43\).

LHS \(= \tan^{-1}\dfrac34 + \tan^{-1}\dfrac43 = \tan^{-1}\dfrac34 + \cot^{-1}\dfrac34 = \dfrac{\pi}{2}\) = RHS (since \(\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}\)).