The correct option is A 1
Substitute x=sinθ,y=sinϕ, so that given equation is reduced to cosθ+cosϕ=a(sinθ−sinϕ)
⇒2cosθ+ϕ2cosθ−ϕ2=2acosθ+ϕ2sinθ−ϕ2
⇒cotθ−ϕ2=a⇒θ−ϕ=2cot−1a
⇒sin−1x−sin−1y=2cot−1a
Differentiating we get 1√1−x2−1√1−y2dydx=0
So the degree is one.