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?
 A: $\arccos \sqrt \frac 1 n = x$ 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 $\tan =\sqrt {\frac 1{n^2} - 1}$
A: Let $\arctan n = a$, then $\tan(a)=n$ and...
$$1 + \tan^2 (a) = 1/(\cos^2 (a) )$$
$$1 + n^2 = 1/ (\cos^2 (a))$$
$$\cos^2 (a) = 1/(1+n^2)$$ 
$$\cos(a) = (1/(1+n^2) )^{1/2}$$
$$\arccos (1/(1+n^2) )^{1/2}   = a = \arctan(n)$$
A: if x=arccos y, than cos x = y, than sin x =sqrt(1-sqr(y)), than tan x = sqrt(1/y² - 1), than x=arctan(1/y² -1);
now set y=sqrt(1/(n²+1)  -> 1/sqr(y) -1 = (n²+1) - 1 = n² , and you have it!
The formula is valid not only for whole numbers, and the functions arctan and arccos has not only one value (only for the result where sin is positive the formula is valid)
