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 $\theta$.
 A: HINT: Formulate a general identity for $ \tan (\pi-x).$ 
A: It’s not particularly efficient, but you can work the problem using very elementary analytic geometry and no trig identities.
Let $P=\left\langle\frac35,\frac45\right\rangle$, $Q=\langle -1,0\rangle$, $R=\langle 1,0\rangle$, and $O=\langle 0,0\rangle$. Clearly angle $POR$ is $\cos^{-1}\frac35$, so angle $POQ$ is $\pi-\cos^{-1}\frac35$. Let $S=\langle x_0,y_0\rangle$ be the point on the unit circle in the upper half plane such that $OS$ bisects angle $POQ$. Then $y_0$ is the perpendicular distance from $S$ to $OQ$, which must be equal to the perpendicular distance from $S$ to $OP$. The latter is the distance from $S$ to the line $y=\frac43x$. 
The perpendicular from $S$ to that line has slope $-\frac34$, so its equation is $y-y_0=-\frac34(x-x_0)$, and it intersects the line $y=\frac43x$ at the point $\langle x_1,y_1\rangle$ where $y_0-\frac34(x_1-x_0)=\frac43x_1$, i.e., where $x_1=\frac{12}{25}\left(y_0+\frac34x_0\right)$ and $y_1=\frac43x_1=\frac{16}{25}\left(y_0+\frac34x_0\right)$. The distance from $\langle x_1,y_1\rangle$ to $S$ is 
$$\begin{align*}&\sqrt{\left(\frac{12}{25}\left(y_0+\frac34x_0\right)-x_0\right)^2+\left(\frac{16}{25}\left(y_0+\frac34x_0\right)-y_0\right)^2}\\
&\qquad\qquad=\sqrt{\left(\frac{12}{25}y_0-\frac{16}{25}x_0\right)^2+\left(\frac{12}{25}x_0-\frac9{25}y_0\right)^2}\\
&\qquad\qquad=\frac1{25}\sqrt{(12y_0-16x_0)^2+(12x_0-9y_0)^2}\\
&\qquad\qquad=\frac1{25}\sqrt{225y_0^2-600x_0y_0+400x_0^2}\\
&\qquad\qquad=\frac15\sqrt{9y_0^2-24x_0y_0+16x_0^2}\\
&\qquad\qquad=\frac15\sqrt{(3y_0-4x_0)^2}\\
&\qquad\qquad=\frac15|3y_0-4x_0|\\
&\qquad\qquad=\frac15(3y_0-4x_0)\;,
\end{align*}$$
since clearly $y_0>0$ and $x_0<0$.
This must be equal to $y_0$, so we have $\frac15(3y_0-4x_0)=y_0$, and hence $y_0=-2x_0$. Thus, the slope of the line $OS$ is $-2$, and it follows that the tangent of angle $SOQ$ is $2$ and hence that $2\tan^{-1}2$ equals angle $POQ$, which we already saw is $\pi-\cos^{-1}\frac35$.
A: $$
2\tan^{-1}(2)
$$
$$
\tan^{-1}\frac{(2*2)}{1-(2*2)}
$$
$$
\tan^{-1}\frac{4}{-3}
$$
$$
\pi-\tan^{-1}\frac{4}{3}
$$
$$
\pi-\cos^{-1}\frac{3}{5}
$$
proved..
A: 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, $\cos(\pi+y)=-\cos y$
