Inverse trigonometry proof

Question

How would I go about proving

$$\arccos x = \arctan \frac{\sqrt {c-x^2}}{x} $$

Where $c$ is a constant and $0< x ≤1$

Answer 1

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 $\tan\theta = \frac {\sin\theta}{\cos\theta} = \frac {\sqrt {1-x^2}}{x}$

$\theta = \arccos x = \arctan \frac {\sqrt {1-x^2}}{x}$

Answer 2

Say $\alpha = \arccos x$ then $\cos \alpha = x$ and $\alpha \in [0,\pi]$ so $$\sin \alpha = \pm \sqrt{1-\cos^2\alpha } = +\sqrt{1-x^2}$$

and thus $$\tan \alpha = {\sin\alpha \over \cos \alpha } = {\sqrt{1-x^2}\over x}$$

so $$\alpha =\arctan {\sqrt{1-x^2}\over x}$$

and thus $$\arccos x =\arctan {\sqrt{1-x^2}\over x}$$