Put \(x = \sin A\), \(y = \sin B\):

\(\cos A + \cos B = a(\sin A - \sin B)\)

\(2\cos\frac{A+B}{2}\cos\frac{A-B}{2} = 2a\cos\frac{A+B}{2}\sin\frac{A-B}{2} \Rightarrow \cot\frac{A-B}{2} = a\)

So \(A - B = \sin^{-1}x - \sin^{-1}y = \) constant. Differentiating:

\(\frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-y^2}}\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \sqrt{\frac{1-y^2}{1-x^2}}\)

The highest derivative \(\frac{dy}{dx}\) appears with power 1, so the degree is 1.

Answer: (A) 1