Demonstrate the height of the cone inscribed in a sphere as a function of the radius of the sphere

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For reference: A cone and a cylinder, both straight, have the same volume and identical bases. Knowing that both are inscribable in a sphere of radius R, what is the height H of the cone (as a function of R)?(A: $\frac{6R}{5}$)

I have not been able to demonstrate this relationship.

$r_{ci}=r_{co} = r\\ r_e=R\\ V_{ci}=V_{co}\implies \pi r^2.h_{ci}=\frac{1}{3}.\pi r^2.h_{ci}\\ \therefore h_{co}=3h_{ci}\\ V_e=\frac{4}{3}.\pi .R^3\\ R^2 = r^2+(h_{co}-R)^2\\ 4R^2=h_{ci}^2+4r^2\implies h_{ci}^2=4(R^2-r^2)\\ (\frac{h_{co}}{3})^2=4(R^2-r^2)\implies h_{co}^2=36(R^2-r^2)\\ h_{co} = 6\sqrt{R^2-r^2}$

I don't see how to continue or even if the question's relationship can exist

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From your work, we subtract the second equation from the first one and then we use the third one: $$\begin{cases} R^2 = r^2+(h_{co}-R)^2\\ R^2=\frac{h_{ci}^2}{4}+r^2\\ h_{co}=3h_{ci} \end{cases}\implies h_{co}-R=\frac{h_{ci}}{2}=\frac{h_{co}}{6} \implies h_{co}=\frac{6R}{5}.$$