\begin{align}
x&=\frac{|AD| \cos\tfrac{\alpha}{2}
+|BE|\cos\tfrac{\beta}{2}
+|CF|\cos\tfrac{\gamma}{2}}{\sin\alpha+\sin\beta+\sin\gamma}
=?\quad (1)
\end{align}

Consider $\triangle ADC$.
We know
that it shares
the circumradius $R=1$
and the side $|AC|$
with $\triangle ABC$,
and we also know all the angles:
$\angle CAD=\tfrac\alpha2$,
$\angle ADC=\beta$ (why?)
and $\angle ACD=\pi-\tfrac\alpha2-\beta$.
Hence, by the law of sines,
\begin{align}
|AD|&=
2R \sin(\pi-\tfrac\alpha2-\beta)
=
2R\sin(\tfrac\alpha2+\beta).
\end{align}
Similarly, from $\triangle BEA$ and $\triangle CFB$,
\begin{align}
|BE|&=
2R\sin(\tfrac\beta2+\gamma),
\\
|CF|&=
2R\sin(\tfrac\gamma2+\alpha).
\end{align}
Numerator of (1) is then
\begin{align}
&|AD|\cos\tfrac\alpha2
+|BE|\cos\tfrac\beta2
+|CF|\cos\tfrac\gamma2
\\
&=2R\left(
\sin(\tfrac\alpha2+\beta)\cos\tfrac\alpha2
+
\sin(\tfrac\beta2+\gamma)\cos\tfrac\beta2
+
\sin(\tfrac\gamma2+\alpha)\cos\tfrac\gamma2
\right)
\\
&=
2R\left(
\tfrac12\sin(\alpha+\beta)+\tfrac12\sin(\beta)
+
\tfrac12\sin(\beta+\gamma)+\tfrac12\sin(\gamma)
+
\tfrac12\sin(\gamma+\alpha)+\tfrac12\sin(\alpha)
\right)
\\
&=
2R\left(
\tfrac12\sin\gamma+\tfrac12\sin\beta
+
\tfrac12\sin\alpha+\tfrac12\sin\gamma
+
\tfrac12\sin\beta+\tfrac12\sin\alpha
\right)
\\
&=
2R(\sin\alpha+\sin\beta+\sin\gamma).
\end{align}
Hence, the answer is $x=2R=2$.