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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def iau2000b(jd_tt):
 """Compute Earth nutation based on the faster IAU 2000B nutation model. `jd_tt` - Terrestrial Time: Julian date float, or NumPy array of floats Returns a tuple ``(delta_psi, delta_epsilon)`` measured in tenths of a micro-arcsecond. Each is either a float, or a NumPy array with the same dimensions as the input argument. The result will not take as long to compute as the full IAU 2000A series, but should still agree with ``iau2000a()`` to within a milliarcsecond between the years 1995 and 2020. """

    dpplan = -0.000135 * 1e7
    deplan =  0.000388 * 1e7

    t = (jd_tt - T0) / 36525.0

    # TODO: can these be replaced with fa0 and f1?

    el  = fmod (485868.249036 +
          t * 1717915923.2178, ASEC360) * ASEC2RAD;

    elp = fmod (1287104.79305 +
             t * 129596581.0481, ASEC360) * ASEC2RAD;

    f   = fmod (335779.526232 +
             t * 1739527262.8478, ASEC360) * ASEC2RAD;

    d   = fmod (1072260.70369 +
             t * 1602961601.2090, ASEC360) * ASEC2RAD;

    om  = fmod (450160.398036 -
             t * 6962890.5431, ASEC360) * ASEC2RAD;

    a = array((el, elp, f, d, om))

    arg = nals_t[:77].dot(a)
    fmod(arg, tau, out=arg)

    sarg = sin(arg)
    carg = cos(arg)

    stsc = array((sarg, t * sarg, carg)).T
    ctcs = array((carg, t * carg, sarg)).T

    dp = tensordot(stsc, lunisolar_longitude_coefficients[:77,])
    de = tensordot(ctcs, lunisolar_obliquity_coefficients[:77,])

    dpsi = dpplan + dp
    deps = deplan + de

    return dpsi, deps