<s>[INST] Geometric identity, cannot show equivalence using trigonometric identities 
clearly 
$$(x+a \cos\theta)^2+(y-a \sin\theta)^2=b^2$$
expanding and using the Weierstrass substitution we find that
$$\theta= 2 \arctan \frac{\left( 2ay- \sqrt{ 4a^2y^2 - ( (x-a)^2+y^2-b^2)( (x+a)^2+y^2-b^2) }\right)}{(x-a)^2+y^2-b^2}  $$
if we use the law of cosines, with $c^2=x^2+y^2$
$$\theta = \arctan_2(y,x) - \arccos( (c^2+a^2-b^2) / (2ac) )$$
Is there a way to pass from one expression to the other using trigonometric identities?
 [/INST] You could look at the angles at the vertex between sides $c$ and $a$. The arctan comes from the $x,y$ triangle and the arccos from the $abc$  one.
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