= Solution
Work on the unit sphere, so side lengths are central angles. Place the right-angle vertex at the north pole and choose perpendicular meridians through it. The other vertices can then be written
$$
B=(\sin a,0,\cos a),\qquad C=(0,\sin b,\cos b).
$$
Their dot product is the cosine of the central angle between them, namely the hypotenuse length $c$. Hence the spherical Pythagorean relation is
$$
\boxed{\cos c=B\cdot C=\cos a\cos b.}
$$
For the small triangle let its hypotenuse be $c_\lambda$. The same formula gives $\cos c_\lambda=\cos(\lambda a)\cos(\lambda b)$, so $c_\lambda\to0$. Expanding each cosine about zero yields
$$
1-\frac12c_\lambda^2+O(c_\lambda^4)
=1-\frac12\lambda^2(a^2+b^2)+O(\lambda^4).
$$
In particular $c_\lambda=O(\lambda)$, and thus $c_\lambda^2=\lambda^2(a^2+b^2)+O(\lambda^4)$. Consequently \b[$(c_\lambda/\lambda)^2\to a^2+b^2$], the ordinary Pythagorean theorem in the locally flat limit.
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