$\Gamma(\dfrac{1}{2}-z)\Gamma(\dfrac{1}{2}+z) = \dfrac{\pi}{cos \pi z}$
thanks for the comment this problem has been solved
$\Gamma(\dfrac{1}{2}-z)\Gamma(\dfrac{1}{2}+z) = \dfrac{\pi}{cos \pi z}$
thanks for the comment this problem has been solved
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Use the well-known Euler's reflection formula, namely,
\begin{equation} \Gamma{(z)}\Gamma{(1 - z)} = \dfrac{\pi}{\sin(\pi z)}. \end{equation}
Taking $z = \frac{1}{2} - z$ yields the equation that you are looking for.
The proof for Euler's reflective equation is given here https://proofwiki.org/wiki/Euler%27s_Reflection_Formula