How to prove the following
$$\,_2F_1\left(-1/2,-1/2,1,k^2 \right)=\frac{2}{\pi}\left(2E+(k^2-1)K \right)$$
where we define
The complete integral of first kind
$$K=K(k) = \int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2x^2}}$$
The complete integral of second kind
$$E=E(k) = \int^1_0 \frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}}\,dx$$
The problem of OP is (presumbaly) that the standard integral representation of the hypergeometric function is not directly applicable for the chosen parameter values. This can be treated as follows.
The integral representation $$_2F_1(a,b;c;t)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1\frac{x^{b-1}(1-x)^{c-b-1}dx}{(1-tx)^a},$$ implies that \begin{align} K&=\frac{\pi}{2}{}_2F_1\left(\frac12,\frac12;1;k^2\right),\\ E&=\frac{\pi}{2}{}_2F_1\left(-\frac12,\frac12;1;k^2\right). \end{align} Therefore the identity we want to prove is equivalent to $$2\, _2F_1\left(-\frac12,\frac12;1;t\right)+(t-1)\,_2F_1\left(\frac12,\frac12;1;t\right)=\, _2F_1\left(-\frac12,-\frac12;1;t\right).$$ This is a special case ($b=-a=\frac12$, $c=1$) of a more general contiguity relation $$(c-a-b)\,_2F_1(a,b;c;t)+a(1-t)\,_2F_1(a+1,b;c;t)=(c-b)\,_2F_1(a,b-1;c;t),$$ which can be proved by comparing the coefficients of different powers of $t$ at both sides.