= Solution
A <branch point> is a point around which analytic continuation changes a value; a branch cut removes curves so continuation becomes single-valued, and a branch is one such single-valued analytic choice. Define
$$
\operatorname{Arcsin}z=\int_0^z(1-t^2)^{-1/2}dt
$$
on $\mathbb C\setminus((-∞,-1]\cup[1,∞))$, choosing the square root equal to $1$ at zero. Monodromy around $\pm1$ explains multivaluedness.
On the upper lip $0+$ of the cut convention, the branch remains positive on $(0,1)$, so $\int_{0+}^1dt/\sqrt{1-t^2}=\pi/2$. Define $\operatorname{Arccos}z=\pi/2-\operatorname{Arcsin}z$ on the same slit domain.
Also
$$
\arctan z=\frac1{2i}[\log(1+iz)-\log(1-iz)].
$$
Changing logarithm branches changes the value by multiples of $\pi$ (already visible at $z=1$). Principal logarithms give a single branch on the plane cut from $\pm i$ outward. Differentiating both sides and checking the value at zero proves
$$
\operatorname{Arctan}z=\operatorname{Arcsin}\frac z{\sqrt{1+z^2}}
$$
where the compatible principal branches are defined.
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