Is $\arctan n$ always equal to $\arccos\sqrt{\frac{1}{n^2+1}}$? $$\arccos\sqrt\frac{1}{2}=\arctan 1$$
$$\arccos\sqrt\frac{1}{5}=\arctan 2$$
$$\arccos\sqrt\frac{1}{10}=\arctan 3$$
$$\arccos\sqrt\frac{1}{17}=\arctan 4$$
$$\arccos\sqrt\frac{1}{26}=\arctan 5$$
$$\arccos\sqrt\frac{1}{37}=\arctan 6$$
$$\arccos\sqrt\frac{1}{50}=\arctan 7$$
The answer is a sequence $n^2+1$ for the slope which is in the inverse of tangent. Digits that are whole numbers. Is there any explanation as to why this is true? Is it a well-known problem?

$\arccos \sqrt \frac 1 n = x$ means $cos x = \frac 1 n$ which means $\sin x = \sqrt{1 - {\sqrt \frac 1 n}^2} = \sqrt{1 - \frac 1 n}$ so $\tan x = \frac{ \sqrt{1 - \frac 1 n}}{\sqrt \frac 1 n} = \sqrt{n - 1}$ so $\arctan{n -1} = x = \arccos \sqrt \frac 1 n$.  So this is known.  
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Another way of looking at it is:
$\tan = \frac {\sin}{\cos} $ and $\sin^2 + \cos^2 = 1$, so $\tan = \frac {\sqrt{1 - \cos^2}}{\cos} = \sqrt {\frac 1{\cos^2} - 1}$ so if $n = \cos x$ then $\tan =\sqrt {\frac 1{n^2} - 1}$
