If $$\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=3$$ then the value of $$\sin^4\theta-\cos^4\theta$$ is 

Given : $$\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=3$$

=> $$sin\theta+cos\theta=3sin\theta-3cos\theta$$

=> $$3sin\theta-sin\theta=3cos\theta+cos\theta$$

=> $$2sin\theta=4cos\theta$$

=> $$\frac{sin\theta}{cos\theta}=\frac{4}{2}$$

=> $$tan\theta=2$$

Using, $$sec^2\theta-tan^2\theta=1$$

=> $$sec^2\theta=1+(2)^2=5$$

$$\therefore$$ $$cos^2\theta=\frac{1}{5}$$

Similarly, $$sin^2\theta=\frac{4}{5}$$

To find : $$\sin^4\theta-\cos^4\theta$$

= $$(sin^2\theta-cos^2\theta)(sin^2\theta+cos^2\theta) = (sin^2\theta-cos^2\theta)$$     [$$\because sin^2\theta+cos^2\theta=1$$]

= $$\frac{4}{5}-\frac{1}{5}=\frac{3}{5}$$

=> Ans - (D)