Rewrite expression involving the hypergeometric function to prove it is real-valued Let
$$
f(\varphi) = \sin(\varphi)^{-i} {}_2F_1(\frac{s - i}{2}, \frac{s - i}{2}, s, \cos^2(\varphi))
$$
where $s \in (1, 2)$ is a real parameter and $\varphi$ is restricted to $(0, \pi/2)$. Prove that $f$ is real-valued.
Background: I had Mathematica solve an ODE in $\varphi$, and this was the output (after some simplification). An argument involving the ODE itself shows that a real-valued solution exists.
I was pleasantly surprised when I plugged in various values of $s$ and $\varphi$ and all the outputs were real-valued. However, I would like to analytically show that this must be real-valued.
I tried various known transformations involving ${}_2F_1, \Gamma,$ and $\sin$, but couldn't get it to work.
Any help is much appreciated!

By http://dlmf.nist.gov/15.8.E1
\begin{align*}
&\sin ^{-i} (\varphi) \times {}_2F_1 \!\left( {\frac{{s - i}}{2},\frac{{s - i}}{2};s;\cos ^2 (\varphi) } \right) \\ & = \sin ^{-i} (\varphi) (1 - \cos ^2 (\varphi) )^{ - \frac{{s - i}}{2}} \times{}_2F_1 \!\left( {\frac{{s - i}}{2},s - \frac{{s - i}}{2};s;\frac{{\cos ^2 (\varphi) }}{{\cos ^2 (\varphi)  - 1}}} \right)
\\ &
 = \sin ^{ - s}(\varphi) \times {}_2F_1 \!\left( {\frac{{s - i}}{2},\frac{{s + i}}{2};s; - \cot ^2 (\varphi) } \right).
\end{align*}
Now
\begin{align*}
& \overline {{}_2F_1\! \left( {\frac{{s - i}}{2},\frac{{s + i}}{2};s; - \cot ^2 (\varphi) } \right)}  = {}_2F_1 \!\left( {\overline {\frac{{s - i}}{2}} ,\overline {\frac{{s + i}}{2}} ;s; - \cot ^2 (\varphi) } \right)
\\ &
 = {}_2F_1 \! \left( {\frac{{s + i}}{2},\frac{{s - i}}{2};s; - \cot ^2 (\varphi) } \right) = {}_2F_1 \!\left( {\frac{{s - i}}{2},\frac{{s + i}}{2};s; - \cot ^2 (\varphi) } \right),
\end{align*}
showing that what multiplies $\sin ^{ - s}(\varphi)$ (which is real) is real.
