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

Given in the question that,

$$\dfrac{sin \theta+cos\theta}{sin \theta-cos\theta}=3$$

Upon cross multiplication, we get

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

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

$$2\cos\theta\ \ =\ \sin\theta\ $$  ---(1)

We know that $$\sin^2\theta\ +\ \cos^2\theta\ \ =\ 1\ $$

Substituting the value of $$\sin\theta\ $$ from (1), we get,

$$4\cos^2\theta\ \ +\ \cos^2\theta\ \ =\ 1$$

$$\cos^2\theta\ \ =\ \dfrac{1}{5}$$

We get, 

$$\sin^2\theta\ \ =\ 1\ -\ \dfrac{1}{5}\ =\ \dfrac{4}{5}$$

The value needed to be calculated is, 

$$\sin^4\theta-\cos^4\theta\ =\ \left(\sin^2\theta\ \right)^2-\ \left(\cos^2\theta\ \right)^2\ =\ \left(\dfrac{4}{5}\right)^2\ -\ \left(\dfrac{1}{5}\right)^2\ =\ \dfrac{16}{25}\ -\ \dfrac{1}{25}\ =\ \dfrac{15}{25}\ =\ \dfrac{3}{5}$$

Hence, the correct answer is option C.