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?

Try this:
$$(x+a \cos\theta)^2+(y-a \sin\theta)^2=b^2$$
 Working through we get $$x\cos\theta-y\sin\theta=\frac {(b^2-a^2-y^2-x^2)}{2a}$$
Now use $x^2+y^2=c^2$ on both sides and set $\phi=\arctan \frac y x$. Divide through by $c$ to obtain:
$$\cos \phi \cos \theta - \sin\phi\sin\theta =\cos(\theta+\phi)=\frac {(b^2-a^2-c^2)}{2ac}$$ and you can do it from there.
