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?

Let $\arctan n = a$, then $\tan(a)=n$ and...
$$1 + \tan^2 (a) = 1/(\cos^2 (a) )$$
$$1 + n^2 = 1/ (\cos^2 (a))$$
$$\cos^2 (a) = 1/(1+n^2)$$ 
$$\cos(a) = (1/(1+n^2) )^{1/2}$$
$$\arccos (1/(1+n^2) )^{1/2}   = a = \arctan(n)$$
