This solves this post and is also related to this MO post by involving $\tfrac12,\tfrac13,\tfrac14,\tfrac16$.

$p=2$

The **singly**-periodic trigonometric functions and the **doubly**-periodic *Jacobi elliptic functions* parameterize the circle so,

$$\sin^2 z+\cos^2 z=1\tag1$$ $$\mathrm{sn}^2(u)+\mathrm{cn}^2(u)=1\tag2$$

If we use these as arguments for hypergeometric ratios, then for $0<z<1$,

$$\frac{_2F_1\Big(\tfrac12,\tfrac12;\tfrac32; \cos^2 z \Big)}{ _2F_1\Big(\tfrac12,\tfrac12;\tfrac32; \sin^2 z \Big)}=\Big(\frac{\pi_\color{blue}2}{2z}-1\Big) \frac{\sin z}{\cos z}\tag3$$

and elliptic integral of the first kind $u = F(\phi,k)$ for $\phi>0$ and *any* $k$,

$$\frac{_2F_1\Big(\tfrac12,\tfrac12;\tfrac32; \mathrm{cn}^2(u) \Big)}{ _2F_1\Big(\tfrac12,\tfrac12;\tfrac32; \mathrm{sn}^2(u) \Big)}=\Big(\frac{\pi_\color{blue}2}{2\phi}-1\Big) \frac{\mathrm{sn}(u) }{\mathrm{cn}(u) }\tag4$$

where $\pi_2 = B\big(\tfrac12,\tfrac12\big)=\pi$. It turns out this can be generalized.

$p=3$

The **doubly**-periodic Dixonian elliptic functions $\mathrm{sm}(z), \mathrm{cm}(z)$ parameterize the *Fermat cubic* so

$$\mathrm{sm}^{3}(z)+\mathrm{cm}^{3}(z)=1\tag5$$

In terms of the Weierstrass elliptic functions,

$$\mathrm{sm}(z)=\frac{-6\wp\left(z;0,\frac1{27}\right)}{3\wp^\prime\left(z;0,\frac1{27}\right)-1},\quad \mathrm{cm}(z)=\frac{3\wp^\prime\left(z;0,\frac1{27}\right)+1}{3\wp^\prime \left(z;0,\frac1{27}\right)-1}$$

After some experimentation with *Mathematica*, we find, for $0<z<1$, the analogous ratio,

$$\frac{_2F_1\Big(\tfrac13,\tfrac23;\tfrac43;\mathrm{cm}^3(z) \Big)}{_2F_1\Big(\tfrac13,\tfrac23;\tfrac43;\mathrm{sm}^3(z) \Big)}=\Big(\frac{\pi_\color{blue}3}{3z}-1\Big) \frac{\mathrm{sm(z)}}{\mathrm{cm(z)}}\tag6$$

where $\pi_3 = B\big(\tfrac13,\tfrac13\big)=\frac{\sqrt3}{2\pi}\Gamma^3\big(\tfrac13\big)$.

Questionfor $p=4,6$

What suitable functions $\alpha(z),\beta(z)$ with, $$\alpha^k(z)+\beta^k(z)=1$$ will solve the hypergeometric ratios below in a similar manner?:

$$\frac{_2F_1\Big(\tfrac14,\tfrac34;\tfrac54;\beta^k(z)\Big)} {_2F_1\Big(\tfrac14,\tfrac34;\tfrac54;\alpha^k(z)\Big)}=\Big(\frac{\pi_{\color{blue}4}}{4z}-1\Big) \frac{\alpha(z)}{\beta(z)}\tag7$$
** or**,

$$\frac{_2F_1\Big(\tfrac16,\tfrac56;\tfrac76;\beta^k(z)\Big)} {_2F_1\Big(\tfrac16,\tfrac56;\tfrac76;\alpha^k(z)\Big)}=\Big(\frac{\pi_{\color{blue}6}}{6z}-1\Big) \frac{\alpha(z)}{\beta(z)}\tag8$$