SOLUTION

Au passage au méridien: $a = H = \delta = 0, z = \phi - \delta$ .

Donc ${dz \over dH} = 0$ et $z = z (H=0) + {1 \over 2} {d^2 z \over dH^2} \Delta H^2$ .

Or ${d^2 z \over dH^2} = \cos a \cos \phi {da \over dH} = \cos a \cos \phi {\cos \delta \cos S \over \sin z}$ .

Donc ${d^2 z \over dH^2} (H=0) = {\cos \delta \cos \phi \over \sin (\phi - \delta)}$ , et finalement: $z = z (H=0) + {1 \over 2} {\cos \delta \cos \phi \over \sin (\phi - \delta)} \Delta H^2$ .