This looked like an interesting academic exercise to work through, even though making the variable substitution increases the amount of effort required. The original region of integration looks like this:

and the integration to be performed works pretty easily over it:
$ \int_0^{\pi/2}\int_0^{\pi/2} \cos\ \theta \ \cos \ \phi \ \ d\theta \ d\phi \ \ = \ \ \int_0^{\pi/2} \sin\ \theta \ \cos \ \phi \ \vert_{\theta = 0}^{\theta = \pi/2} \ \ d\phi \ $
$ = \ \ \int_0^{\pi/2} \ ( \ 1 \ - \ 0 \ ) \ \cos \ \phi \ \ \ d\phi \ \ = \ \ \sin \ \phi \ \vert_0^{\pi/2} \ = \ 1 \ - \ 0 \ = \ 1 \ \ . $
(I've carried out this calculation for comparison against the alternate method.)
Under the variable substitution PanAkry proposes, $ \ \alpha \ = \ \theta+\phi \ \ $ and $ \ \ \beta \ = \ \theta-\phi \ $ , we can transform the lines along which the sides of the original square lie in the $ \ \theta - \phi \ $ plane to obtain
$ \mathbf{\overline{AB}} \ : \quad \phi \ = \ 0 \ \ \rightarrow \ \ \alpha \ = \ \theta \ + \ 0 \ \ , \ \ \beta \ = \ \theta \ - \ 0 \ \ \Rightarrow \ \ \beta \ = \ \alpha \ \ ; $
$ \mathbf{\overline{BC}} \ : \quad \theta \ = \ \frac{\pi}{2} \ \ \rightarrow \ \ \alpha \ = \ \frac{\pi}{2} \ + \ \phi \ \ , \ \ \beta \ = \ \frac{\pi}{2} \ - \ \phi \ \ \Rightarrow \ \ \beta \ = \ \pi \ - \ \alpha \ \ ; $
$ \mathbf{\overline{CD}} \ : \quad \phi \ = \ \frac{\pi}{2} \ \ \rightarrow \ \ \alpha \ = \ \theta \ + \ \frac{\pi}{2} \ \ , \ \ \beta \ = \ \theta \ - \ \frac{\pi}{2} \ \ \Rightarrow \ \ \beta \ = \ \alpha \ - \pi \ ; $
$ \mathbf{\overline{DA}} \ : \quad \theta \ = \ 0 \ \ \rightarrow \ \ \alpha \ = \ 0 \ + \ \phi \ \ , \ \ \beta \ = \ 0 \ - \ \phi \ \ \Rightarrow \ \ \beta \ = \ -\alpha \ \ . $
We find that the sides of the transformed square in the $ \ \alpha - \beta \ $ plane indeed fall along straight lines; solving for the intersections of these lines gives the four transformed vertices described by Michael Hardy in his answer. The graph for this region is shown below.

The transformation has the effect of changing the area of the square (which is what its non-unit Jacobian tells us) and also rotates it, which complicates the integration slightly, as we must divide the region in half (this will be the case regardless of the order of integration):
$ \rightarrow \ \ \int_0^{\pi / 2} \int_{-\alpha}^{\alpha} \ \frac{1}{2} \ ( \ \cos \alpha \ + \ \cos \beta \ ) \ \vert \ \mathfrak{J} \ \vert \ \ d\beta \ d\alpha $ $ \ \ + \ \ \int_{\pi / 2}^{\pi} \ \int_{\alpha - \pi}^{\pi - \alpha} \ \frac{1}{2} \ ( \ \cos \alpha \ + \ \cos \beta \ ) \ \vert \ \mathfrak{J} \ \vert\ \ d\beta \ d\alpha \ \ . $
We can compute the Jacobian from (for example)
$ \mathfrak{J}^{-1} \ \ = \ \ \left[\begin{array}{cc}\frac{\partial \alpha}{\partial \theta}&\frac{\partial \alpha}{\partial \phi}\\\frac{\partial \beta}{\partial \theta}&\frac{\partial \beta}{\partial \phi}\end{array}\right] \ \ = \ \ \left[\begin{array}{cc}1& \ 1\\1&-1\end{array}\right] \ \ , \ \ \det \ \mathfrak{J}^{-1} \ = \ -2 $
$ \mathfrak{J} \ \ = \ \ \frac{1}{\det \ \mathfrak{J}^{-1}} \ \left[\begin{array}{cc}-1& \ -1\\-1& \ 1\end{array}\right] \ \ = \ \ \left[\begin{array}{cc}\frac{1}{2}& \ \frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}\end{array}\right] \ \ \Rightarrow \ \ \det \ \mathfrak{J} \ = \ -\frac{1}{2} \ \ . $
(It may be noted that since the determinant of the Jacobian is negative, the orientation of the transformed square is reversed: the circulation around the transformed vertices $ \ A \ 'B \ 'C \ 'D \ ' \ $ is now clockwise.)
Carrying out the integration produces
$ \frac{1}{2} \cdot \frac{1}{2} \ \int_0^{\pi / 2} \int_{-\alpha}^{\alpha} \ ( \ \cos \alpha \ + \ \cos \beta \ ) \ \ d\beta \ d\alpha $ $\ \ + \ \ \frac{1}{2} \cdot \frac{1}{2} \ \int_{\pi / 2}^{\pi} \ \int_{\alpha - \pi}^{\pi - \alpha} \ ( \ \cos \alpha \ + \ \cos \beta \ ) \ \ d\beta \ d\alpha \ \ $
$ = \ \ \frac{1}{4} \ \left[ \ \int_0^{\pi / 2} \ ( \ \beta \ \cos \alpha \ + \ \sin \beta \ ) \ \vert_{-\alpha}^{\alpha} \ \ d\alpha \ \ + \ \ \int_{\pi / 2}^{\pi} \ ( \ \beta \ \cos \alpha \ + \ \sin \beta \ ) \ \vert_{\alpha - \pi}^{\pi - \alpha} \ \ d\alpha \ \right] \ $
$ = \ \ \frac{1}{4} \ \left[ \ \int_0^{\pi / 2} \ 2 \ ( \ \alpha \ \cos \alpha \ + \ \sin \alpha \ ) \ \ d\alpha \ \ + \ \ \int_{\pi / 2}^{\pi} \ 2 \ ( \ [\pi - \alpha] \ \cos \alpha \ + \ \sin [\pi - \alpha] \ ) \ \ d\alpha \ \right] $
$ = \ \ \frac{1}{2} \ \left[ \ ( \ \alpha \ \sin \alpha \ + \ \cos \alpha \ + \ [- \cos \alpha ] \ ) \ \vert_0^{\pi / 2} \quad + \ ( \ \pi \ \sin \alpha \ - \ [ \ \alpha \ \sin \alpha \ + \ \cos \alpha \ ] \ + \ [ -\cos \alpha ] \ ) \ \vert_{\pi / 2}^{\pi} \ \right] $
$ = \ \ \frac{1}{2} \ \left[ \ ( \ \alpha \ \sin \alpha \ ) \ \vert_0^{\pi / 2} \quad + \ ( \ \pi \ \sin \alpha \ - \ \alpha \ \sin \alpha \ - \ 2 \ \cos \alpha \ ] \ ) \ \vert_{\pi / 2}^{\pi} \ \right] $
$ = \ \ \frac{1}{2} \ \left[ \ \left( \ \frac{\pi}{2} \ \sin \frac{\pi}{2} \ - \ 0 \ \right) \ + \ \left( \ 0 \ - \ 0 \ - \ 2 \ \cos \pi \ - \ \pi \ \sin \frac{\pi}{2} \ + \ \frac{\pi}{2} \ \sin \frac{\pi}{2} \ + \ 0 \ \right) \ \right] $
$ = \ \ \frac{1}{2} \ \left[ \ \frac{\pi}{2} \cdot 1 \ - \ 2 \ (-1) \ - \ \pi \cdot 1 \ + \ \frac{\pi}{2} \cdot 1 \ \right] \ = \ 1 \ \ . $
So we recover the value of the original integral, having made use of a "product-to-sum" trigonometric identity, albeit with somewhat more struggle (I wouldn't recommend taking this route on an exam...).