How to prove Trigonometry equation? how to solve following equation 
$$
\tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{1}{9}\right)
=
\cos^{-1}\left(\frac{3}{5}\right)
$$
How to prove the above equation?
 A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\tan\pars{\overbrace{\cos^{-1}\left(\frac{3}{5}\right)}^{\ds{\equiv\ x}}}
=
\tan\pars{\tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{1}{9}\right)}
=
{1/4 + 1/9 \over 1 - \pars{1/4}\pars{1/9}} = {13 \over 35}
\end{align}
$$
\tan\pars{x} = {\root{1 - \cos^{2}\pars{x}} \over \cos\pars{x}}
=
{\root{1 - \pars{3/5}^{2}} \over 3/5} = {4/5 \over 3/5} = {4 \over 3}
\color{#0000ff}{\Huge\not=} {13 \over 35}
$$
A: From this or Ex$\#5$  of Page $\#276$ of this  $$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$$ if $xy<1$
Now, as the principal value of $\tan$ lies $\in\left[-\frac\pi2,\frac\pi2\right],$
If $\displaystyle \tan^{-1}z=\theta,\tan\theta=z,$
$\displaystyle\cos\theta=+\frac1{\sqrt{1+z^2}}$
A: After you apply the formula, let $cos \theta= \frac{3}{5}$. Now convert this to $tan \theta$, which is trivial, and compare both sides.
