Trigonometry equation problem Show that
$$\arctan2 = \arccos(1/\sqrt5).$$
This is the result of solving one trigonometrical equation.

Let $\theta =\arctan 2$. Then $\tan \theta =2$ and we need to prove that $\tan \theta=\tan (\arccos (1/\sqrt{5}))=2$. One way is to use the identity
\begin{equation*}
\tan x=\pm  \sqrt{\frac{1}{\cos ^{2}x}-1},
\end{equation*}
which can be derived from the fundamental identity
\begin{equation*}
\cos ^{2}x+\sin ^{2}x=1.
\end{equation*}
For $x=1/\sqrt{5}$ we have that
\begin{eqnarray*}
\tan (\arccos (1/\sqrt{5})) &=&\sqrt{\frac{1}{\cos ^{2}(\arccos (1/\sqrt{5}))}
-1} \\ && \\
\text{(The sign is positive because }0 &<&\arccos (1/\sqrt{5})<\pi /2\text{)}
\\ &&\\
&=&\sqrt{\frac{1}{(1/\sqrt{5})^{2}}-1}=2.
\end{eqnarray*}
