On the Wolfram MathWorld page of Legendre Differential Equation, Legendre polynomials are represented as
$$ P_l(x) = c_n \begin{cases}\begin{align*} &_2F_1\left(-\frac{1}{2}(l), \frac{1}{2}(l + 1); \frac{1}{2}; x^2\right) && \text{for even } l\\ &x_2F_1\left(\frac{1}{2}(l + 2), \frac{1}{2}(1 - l); \frac{3}{2}; x^2\right) && \text{for odd } l \\ \end{align*}\end{cases}\tag{1} $$
I also know Legendre polynomials can be written in this single hypergeometric function.
$$ {}_2F_1\left(l + 1, -l; 1; \frac{1 - x}{2}\right) \tag{2} $$
How to merge (1) to get (2)?
A quadratic transform for the Gaussian hypergeometric function exists which links (1) and (2): \begin{align} { }_2 F_1\left(a, b ; \frac{a+b+1}{2} ; z\right)&= \\ &\frac{\sqrt{\pi} \Gamma\left(\frac{a+b+1}{2}\right)}{\Gamma\left(\frac{a+1}{2}\right) \Gamma\left(\frac{b+1}{2}\right)}{ }_2 F_1\left(\frac{a}{2}, \frac{b}{2} ; \frac{1}{2} ;(2 z-1)^2\right)+\\ &+\frac{2 \sqrt{\pi}(2 z-1) \Gamma\left(\frac{a+b+1}{2}\right)}{\Gamma\left(\frac{a}{2}\right) \Gamma\left(\frac{b}{2}\right)}{ }_2 F_1\left(\frac{a+1}{2}, \frac{b+1}{2} ; \frac{3}{2} ;(2 z-1)^2\right) \end{align} Here, by choosing $a=-l,b=l+1,z=(1-x)/2$ \begin{align} {}_2F_1\left( -l,l+1; 1; \frac{1 - x}{2}\right)&=\frac{\sqrt{\pi}}{\Gamma\left((1-l)/2\right) \Gamma\left(1+l/2\right)}{ }_2 F_1\left(-l/2, (1+l)/2 ;1/2 ;x^2\right)+\\ &+2x\frac{ \sqrt{\pi}}{\Gamma\left(-l/2\right) \Gamma\left((1+l)/2\right)}{ }_2 F_1\left((1-l)/2, 1+l/2 ; 3/2 ;x^2\right) \end{align} Now, due to the $\Gamma\left((1-l)/2\right)$ and $\Gamma\left(-l/2\right)$ in the denominators of the coefficients, the second (the first) term vanishes when $l$ is even (odd).