If $$x = a \sec q + b \tan q$$  and $$y = a \tan q + b \sec q$$ (a ≠ b), then the value of $$(x^2 - y^2)/a^2 - b^2)$$ is

$$x = a \sec q + b \tan q$$

$$=$$>  $$x^2=\left(a\sec q+b\tan q\right)^2$$

$$=$$>  $$x^2=a^2\sec^2q+b^2\tan^2q+2ab\sec q\tan q$$

$$y = a \tan q + b \sec q$$

$$=$$>  $$y^2=\left(a\tan q+b\sec q\right)^2$$

$$=$$>  $$y^2=a^2\tan^2q+b^2\sec^2q+2ab\sec q\tan q$$

$$\therefore\ x^2-y^2=a^2\sec^2q-a^2\tan^2q+b^2\tan^2q-b^2\sec^2q+2ab\sec q\tan q-2ab\sec q\tan q$$

$$=$$>  $$x^2-y^2=a^2\left(\sec^2q-\tan^2q\right)-b^2\left(\sec^2q-\tan^2q\right)$$

$$=$$>  $$x^2-y^2=a^2\left(1\right)-b^2\left(1\right)$$

$$=$$>  $$x^2-y^2=a^2-b^2$$

$$=$$>  $$\frac{x^2-y^2}{a^2-b^2}=1$$

Hence, the correct answer is Option A