<s>[INST] Inverse trigonometry proof How would I go about proving
$$\arccos x  = \arctan \frac{\sqrt {c-x^2}}{x} $$
Where $c$ is a constant and $0< x ≤1$
 [/INST] Suppose $\theta = \arccos x$ 
then $\cos \theta = x$
if $0<x\le 1$ then $0\le \theta < \frac {\pi}{2}$
$\sin \theta = \sqrt {1-x^2}$ and $\tan\theta = \frac {\sin\theta}{\cos\theta} =  \frac {\sqrt {1-x^2}}{x}$
$\theta = \arccos x = \arctan \frac {\sqrt {1-x^2}}{x}$
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