If $$x=a(sin\ \theta + cos\ \theta)$$ and $$y=b(sin\ \theta - cos\ \theta)$$, then the value of $$\frac{x^{2}}{a^{2}}+\frac{v^{2}}{b^{2}}$$ is:

Given : $$x=a(sin\ \theta + cos\ \theta)$$ and $$y=b(sin\ \theta - cos\ \theta)$$

Squaring both sides, we get :

=> $$x^2=a^2(sin\ \theta+cos\ \theta)^2$$

=> $$x^2=a^2(sin^2\ \theta+cos^2\ \theta+2sin\ \theta.cos\ \theta)$$

=> $$x^2=a^2(1+2sin\ \theta.cos\ \theta)$$

=> $$\frac{x^2}{a^2}=1+2sin\ \theta.cos\ \theta$$ ----------------(i)

Similarly, $$\frac{y^2}{b^2}=1-2sin\ \theta.cos\ \theta$$ ----------------(i)

Adding both equations (i) and (ii), 

=> $$\frac{x^{2}}{a^{2}}+\frac{v^{2}}{b^{2}}$$ $$=(1+2sin\ \theta.cos\ \theta)+(1-2sin\ \theta.cos\ \theta)$$

= $$1+1=2$$

=> Ans - (C)