If $$x = a \sec \theta + b \tan \theta  and  y = a \tan \theta + b \sec \theta$$, then find $$x^2 - y^2$$.

$$x=a\sec\theta\ +b\tan\theta\ \ .$$

or,$$x^2=\left(a\sec\theta\ +b\tan\theta\right)^2=a^2\sec^2\theta\ +b^2\tan^2\theta\ +2ab\sec\theta\ \tan\theta\ .$$

$$y^2=\left(a\tan\theta\ +b\sec\theta\right)^2=a^2\tan^2\theta\ +b^2\sec^2\theta\ +2ab\sec\theta\ \tan\theta\ .$$

So,$$x^2-y^2=a^2\sec^2\theta\ +b^2\tan^2\theta\ +2ab\sec\theta\ \tan\theta\ -a^2\tan^2\theta\ -b^2\sec\theta\ -2ab\sec\theta\ \tan\theta\ .$$

or,$$x^2-y^2=a^2\left(\sec^2\theta-\tan^2\theta\ \right)\ -b^2\left(\sec^2\theta\ -\tan^2\theta\ \right)=a^2-b^2.\ \ \ \left(as\ ,\ \sec^2\theta\ -\tan^2\theta\ =1\right).$$

A is correct choice.