Proving $\frac { \Gamma(1-s) \Gamma(s) } {\Gamma(s/2) } = \frac{ 2^{s-1} \sqrt{\pi}} {\cos(\pi s/2) } $ I am trying a question in complex analysis and there solution was given but I am having trouble proving an identity involving Gamma Function.

Prove: $$\frac { \Gamma(1-s) \Gamma(s) } {\Gamma(s/2) } = \frac{ 2^{s-1} \sqrt{\pi}} {\cos(\pi s/2) } $$

I tried using Formula $\Gamma(s) =s\Gamma(s) $ and $\Gamma(s)\Gamma(1-s) = \frac{\pi}{ \sin(\pi s)}$, but in vain.
So, can you please tell how to prove them and which results you are using?
 A: This cannot be true.
Consider
$$f(s)=\frac { \Gamma(1-s) \Gamma(s) } {\Gamma \left(\frac{s}{2}\right) }= \frac{\pi  \csc (\pi  s)}{\Gamma \left(\frac{s}{2}\right)} \qquad \text{and} \quad g(s)=\sqrt{\pi } \,2^{s-1} \sec \left(\frac{\pi  s}{2}\right)$$
Using Taylor series
$$f(s)=\frac{1}{2}+\frac{\gamma  }{4}s+O\left(s^2\right)\qquad \text{and} \quad g(s)=\frac{\sqrt{\pi }}{2}+\frac{\sqrt{\pi }  \log (2)}{2} s+O\left(s^2\right)$$
Edit (after @Raymond Manzoni's answer)
If I had been less lazy, suspecting a typo in the definition of $f(s)$, Taylor series would have given the answer.
Suppose that
$$f(s)=\frac {\Gamma \left(\sum _{i=0}^n a_i \,s^i \right) \Gamma(s) } {\Gamma \left(\frac{s}{2}\right) }$$
Expanding $\big[f(s)-g(s)\big]$ as a Taylor series around $s=0$, we would have obtained
$$\big[f(s)-g(s)\big]=\sum _{i=0}^n b_i \,s^i$$ and the successive results
$$b_0=\frac{1}{2} \left(\Gamma (a_0)-\sqrt{\pi }\right)=0 \implies a_0=\frac 12$$
$$b_1=-\frac{1}{4} \sqrt{\pi } (2 a_1+1) (\gamma +2\log (2))=0 \implies a_1=-\frac 12$$
$$b_2=-\frac{1}{2} \sqrt{\pi } a_2 (\gamma +2\log (2))=0 \implies a_2=0$$
$$b_3=-\frac{1}{2} \sqrt{\pi } a_3 (\gamma +2\log (2))=0\implies a_3=0$$
$$b_4=-\frac{1}{2} \sqrt{\pi } a_4 (\gamma +2\log (2))=0\implies a_4=0$$  and so on.
A: A third very useful identity is the duplication formula $(18)$ :
$$\tag{1}\Gamma(2z)=(2\pi)^{-1/2}\,{2^{\,2z-1/2}}\;\Gamma(z)\,\Gamma\left(z+\frac 12\right)$$
\begin{align}
&\text{For $\,z=\dfrac s2\,$ this becomes}\\
\Gamma(s)&=(2\pi)^{-1/2}\,{2^{\,s-1/2}}\,\Gamma\left(\frac s2\right)\,\Gamma\left(\frac {s+1}2\right)\\
&\text{let's multiply by $\Gamma\left(\small{\frac {1-s}2}\right)$}\\
\tag{2}\Gamma(s)\;\Gamma\left(\frac {1-s}2\right)&=(2\pi)^{-1/2}\,{2^{\,s-1/2}}\,\Gamma\left(\frac s2\right)\,\Gamma\left(\frac {s+1}2\right)\Gamma\left(\frac {1-s}2\right)\\
&\text{and use the reflection formula :}\\
\tag{3}\Gamma\left(\frac {s+1}2\right)\Gamma\left(1-\frac {s+1}2\right) &= \frac{\pi}{ \sin\left(\pi \frac {s+1}2\right)}= \frac{\pi}{ \cos\left(\pi \frac s2\right)}\\
&\text{combining $(2)$ and $(3)$  gives :}
\end{align}
$$\tag{4}\boxed{\frac{\Gamma(s)\,\Gamma\left(\frac{1-s}2\right)}{\Gamma\left(\frac s2\right)}=\frac{2^{\,s-1}\,\sqrt{\pi}}{ \cos\left(\pi \frac s2\right)}}$$
There was merely a division by $2$ missing in one of the gamma functions!
