If $$\tan \theta = \frac{a}{b}$$ such that $$\frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta} = (\frac{a^2 - b^2}{a^2 + b^2})^k$$, then k is?

$$\frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta}$$, divided by $$\cos\ \theta\ $$ in numinator and denominator,

then it becomes $$\ \frac{\ a\tan\ \theta\ -b}{a\tan\ \theta\ +b}$$, from question we know the value of tan$$\theta\ $$=$$\ \frac{\ a}{b}$$

$$\ \ \ \frac{\ a\times\ \ \frac{\ a}{b}-b}{a\times\ \ \frac{\ a}{b}+b}$$=$$\ \frac{\ a^2-b^2}{a^2+b^2}$$

so k =1 

option A has value 1,

Hence option 'A' is correct.