Show that $2\tan^{-1}(2) = \pi - \cos^{-1}(\frac{3}{5})$ 
Show that $2\tan^{-1}(2) = \pi - \cos^{-1}(\frac{3}{5})$

So, taking $\tan$ of both sides I get:
LHS $=\frac{2\tan(\tan^{-1}(2))}{1 - \tan^2(\tan^{-1}(2))} = -\frac{4}{3}$
and
RHS $= \tan(\pi - \cos^{-1}(\frac{3}{5})) = ~...$
I wasn't sure how to treat the RHS given the $\pi$ term. If it were just $\cos^{-1}(\frac{3}{5})$ I guess I would try to manipulate $\sin^2(\theta) + \cos^2(\theta) = 1$ to try and find something appropriate for substitution; however, I am not sure where to even start with the given $\theta$.

From this or Ex$\#5$  of Page $\#276$ of this,
  $$\tan^{-1}x+\tan^{-1}y=\begin{cases} \tan^{-1}\frac{x+y}{1-xy} &\mbox{if } xy<1 \\\pi+ \tan^{-1}\frac{x+y}{1-xy} & \mbox{if } xy> 1. \end{cases} $$ 
$$\implies2\tan^{-1}x=\begin{cases} \tan^{-1}\frac{2x}{1-x^2} &\mbox{if } x^2<1\iff -1\le x\le1 \\\pi+ \tan^{-1}\frac{2x}{1-x^2} & \mbox{ else where } \end{cases} $$ 
$$\implies2\tan^{-1}2=\tan^{-1}\left(\frac{2\cdot2}{1-2^2}\right)=\pi+\tan^{-1}\left(-\frac43\right) $$
Let $\theta=\tan^{-1}\left(-\frac43\right)\implies \tan\theta=-\frac43$
Now, as the principal value of $\displaystyle\tan$ lies $\displaystyle\in\left[-\frac\pi2,\frac\pi2\right],\cos\theta=+\frac1{\sqrt{1+\tan^2\theta}}$
Now, $\cos(\pi+y)=-\cos y$
