During solving some spherical problem by 2 independent ways I found "two" solutions represented by $f(t)$
$$f_1(t)=\frac{\sqrt{1+t}(2-t)}{t(\sqrt{1+t}-\sqrt{1-t)}},$$
and
$$f_2(t) =\frac{(t-2)(1-t^{2}+\sqrt{1-t^{2}})}{t^{2}(t-1-\sqrt{1-t^{2}})},$$
where $t\in(0,1)$. Suprisingly, both formulas seem to be identical for any $t\in(0,1)$, which I found challenging. For example, $f_1(0.5)=f_2(0.5)=7.0981$. Hence, I tried to verify their identity
$$ \begin{align*} f_{2}(t) & =\frac{(2-t)\left[(1-t)(1+t)+\sqrt{1-t}\sqrt{1+t}\right]}{t^{2}(\sqrt{1-t}\sqrt{1-t}+\sqrt{1-t}\sqrt{1+t})},\\ & =\frac{(2-t)\left[\sqrt{1-t}\sqrt{1-t}(1+t)+\sqrt{1-t}\sqrt{1+t}\right]}{t^{2}(\sqrt{1-t}\sqrt{1-t}+\sqrt{1-t}\sqrt{1+t})},\\ & =\frac{(2-t)\left[\sqrt{1-t}(1+t)+\sqrt{1+t}\right]}{t^{2}(\sqrt{1-t}+\sqrt{1+t})}, \end{align*} $$
but my attempts after few steps finished :-).
Could I kindly ask for your help? Thank you very much...
Updated proof:
Multiplying $f_{1}(t)$ by $\sqrt{1+t}+\sqrt{1-t}$ leads to: $$ \begin{align*} f_{1}(t) & =\frac{(2-t)\left[1+t+\sqrt{1+t}\sqrt{1-t}\right]}{2t^{2}},\\ & =\frac{(2-t)(1+t+\sqrt{1-t^{2}})}{2t^{2}}. \end{align*}$$
Multiplying $f_{2}(t)$ by $t-1+\sqrt{1-t^{2}}$ leads to:
$$ \begin{align*} f_{2}(t) & =\frac{(t-2)\left[t-t^{3}+t\sqrt{1-t^{2}}-t^{2}\sqrt{1-t^{2}}\right],}{2t^{3}(t-1)}\\ & =\frac{(2-t)\left[t^{2}-1+\sqrt{1-t^{2}}(t-1)\right]}{2t^{2}(t-1)},\\ & =\frac{(2-t)\left[(t-1)(t+1)+\sqrt{1-t^{2}}(t-1)\right]}{2t^{2}(t-1)},\\ & =\frac{(2-t)(1+t+\sqrt{1-t^{2}})}{2t^{2}}. \end{align*} $$
Hence, $f_1(t)=f_2(t).$
Hint: Multiply numerator and denominator of $$f_1(t)=\frac{\sqrt{1+t}(2-t)}{t(\sqrt{1+t}-\sqrt{1-t})}$$ by $$\sqrt{1+t}+\sqrt{1-t}$$ and multiply numerator and denominator of $$f_2(t)=\frac{(t-2)(1-t^2+\sqrt{1-t^2})}{t^2(t-1-\sqrt{1-t^2})}$$ by $$1-t+\sqrt{1-t^2}$$