Finding the northernmost latitude in a great circle that passes through two points on the sphere I'm trying to solve the following problem from Smart's Text-Book on Spherical Astronomy (exercise 5 on p.23 of the 6th ed.):

$A$ and $B$ are two places on the earth's surface with the same latitude $\phi$; the difference of longitude between $A$ and $B$ is $2l$. Prove that

*

*the highest latitude reached by the great circle $AB$ is $\tan^{-1} ( \tan \phi \sec l )$, and

*the distance measured along the parallel of latitude between $A$ and $B$ exceeds the great circle distance $AB$ by $$2 \csc 1' [ l \cos \phi - \sin^{-1} ( \sin l \cos \phi ) ]\text{ nautical miles}.$$

I have tried everything I can think of to solve the first part, and I fell like I'm missing something obvious. I try to solve the triangle, use trigonometric identities, sine law, polar triangles, and nothing works.
 A: I think it is a little more straightforward if you work in Cartesian coordinates.
Take an earth of radius one, for simplicity.
Then we can take the longitudes as $\pm l$ without loss of generality. Then
the halfway point between the two points, projected to the surface of the earth, will have the highest latitude (equivalently the highest $z$ component).
The two points are $(\cos \phi \cos l, \cos \phi \sin l, \sin \phi)$ and
$(\cos \phi \cos l, -\cos \phi \sin l, \sin \phi)$, and the mid point is
$(\cos \phi \cos l, 0, \sin \phi)$. To project to the surface, we divide by
the norm, to get a $z$ component (on our unit earth) of
$\sin \delta = {\sin \phi \over \sqrt{ (\cos \phi \cos l)^2 + \sin^2 \phi}} = { \tan \phi \over \sqrt{\cos^2l+\tan^2 \phi} }$.
To obtain the $\arctan$, we first need $\cos \delta$, which we get from
$\cos \delta = \sqrt{1 -\sin^2 \delta} =  { \cos l \over \sqrt{ \cos^2l+\tan^2 \phi} } $, and hence
$\tan \delta = {\tan \phi \over \cos l}$, which is the desired result.
A: Let $C$ be the point of highest latitude on the great circle $AB$,
and let $N$ be the north pole.
Then you have a spherical triangle $ACN$ with a right angle at $C$
and the angle $\ell$ at $N$, and the arc length of $AN$ is 
$\frac\pi2 - \varphi.$
From the spherical law of sines you can get the arc length of $AC$.
Now you have the arc lengths of two sides of the triangle and you
have the angles opposite each of those two sides;
look for a formula that lets you calculate the arc length of the third side
of the spherical triangle (that is, the side $AC$)
from that given information.
EDITED You might try Napier's Analogies.
(It looks like quite a bit of additional manipulation might be
needed to arrive at the desired formula.)
Note that as you get the length of $AC$ as part of the procedure above,
and the distance along the line of latitude is easily gotten
from $\ell$ and $\varphi$, the second part of the
problem should come relatively easily.
EDITED Rather than Napier, try the cotangent formulas,
in particular,
$$\cos a \cos B = \cot c \sin a - \cot C \sin B$$
(but note that this is quoting the formula for the spherical triangle
$ABC$, whereas you are working with $ANC$, so instead of $B$
you want the angle at $N$, namely $\ell$; also, the $c$ in this formula
is $\frac\pi2 - \varphi$ in your notation, and $a$ is $\frac\pi2$ minus
the latitude of $C$, which you are trying to find).
