What is metric of spherical coordinates $(t,r,\theta,\phi)$? In spherical coordinates the flat space-time metric takes:
$$ds^2=-c^2dt^2+dr^2+r^2d\Omega^2$$
where $r^2d\Omega^2$ come from when the signature of metric $g_{\mu\nu}$ is  (-,+,+,+)?
what is signature of spherical metric?

this is signature of spherical coordinates:
$(1,r^2,r^2\sin^2\theta$)

let me ask in this way,
Isn't it metric of spherical coordinates?
$$g_{\mu\nu}= (-1,+1,+r^2,+r^2\sin^2\theta)$$
 A: Metric signature is a coordinate-invariant notion.  Given a metric, one computes the number of positive and negative eigenvalues that it has, and this gives its signature.  For a diagonal metric, like the metric
$$
  ds^2 = dr^2 + r^2 d\theta^2
$$
both diagonal components are positive, so the metric has precisely two positive eigenvalues, and its signature is therefore $(+,+)$.
A: The trickiness is what you mean by a spherical metric.  What you've written down is the metric of flat space in spherical coordinates, which can be thought of as a warped product of the flat minkowskian two space $(t,r)$ with the unit sphere.  This space is equivalent to the normal $(t,x,y,z)$ coordinates of standard special relativity under a coordinate transformation.
There are other choices, however.  In particular, you can have a space where the constant-time hypersurfaces are 3-spheres, rather than 2-spheres.  Here, the metric will be:
$$ds^{2} = -dt^{2} + d\psi^{2} + \sin^{2}\psi d\theta^{2} + sin^{2}\psi\sin^{2}\theta d\phi^{2}$$
You will find that this space is NOT equivalent to flat space.  There is other trickery you can do with spheres to get yet more inequivalent spaces.
