Step 1: Set A = 10 degrees and B = 50 degrees:
Then $A+B=60^{\circ}$ and $A-B=-40^{\circ}$. We use the identity $\cos^2A+\cos^2B=1+\cos(A+B)\cos(A-B)$.
Step 2: Evaluate the squares:
$\cos^2A+\cos^2B=1+\cos60^{\circ}\cos40^{\circ}=1+\tfrac12\cos40^{\circ}$.
Step 3: Handle the product and finish:
$\cos A\cos B=\tfrac12\cos60^{\circ}+\tfrac12\cos40^{\circ}=\tfrac14+\tfrac12\cos40^{\circ}$. The difference is $1-\tfrac14=\tfrac34$, option C.
Final Answer:
The cos 40 degree terms cancel and 3/4 remains.
\[ \boxed{\text{(C) }\dfrac34} \]