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Solution
We have cos 7A +Cos 5A = 2cos 6A cos A (1) cos C +cos D = 2cos(C+D)/2 cos (C-D)/2
Now
sin7A -sin5A = 2cos6AsinA (2) sinC -sinD = 2cos(C+D)/2 sin(C-D)/2
Dividing (1) and (2)
we get $$\frac{\cos7A+\cos5A}{\sin7A-\sin5A}=\frac{\cos A}{\sin A}=\ \cot A$$
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