\[ \frac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta} = 3 \] Or, \[ \frac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta} = \frac{3}{1} \] Using componendo and dividendo: \[ \frac{(\sin \theta + \cos \theta) + (\sin \theta - \cos \theta)}{(\sin \theta + \cos \theta) - (\sin \theta - \cos \theta)} = \frac{3 + 1}{3 - 1} \] \[ \frac{\sin \theta + \cos \theta + \sin \theta - \cos \theta}{\sin \theta + \cos \theta - \sin \theta + \cos \theta} = \frac{4}{2} \] \[ \frac{2\sin \theta}{2\cos \theta} = \frac{2}{1} \] Squaring both sides: \[ \left(\frac{\sin \theta}{\cos \theta}\right)^2 = \left(\frac{2}{1}\right)^2 \Rightarrow \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{4}{1} \] Now applying componendo and dividendo again: \[ \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta - \cos^2 \theta} = \frac{4 + 1}{4 - 1} = \frac{5}{3} \] Since \(\sin^2 \theta + \cos^2 \theta = 1\), \[ \frac{1}{\sin^2 \theta - \cos^2 \theta} = \frac{5}{3} \Rightarrow \sin^2 \theta - \cos^2 \theta = \frac{3}{5} \] Multiplying both sides by 1: \[ (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) = \frac{3}{5} \Rightarrow \sin^4 \theta - \cos^4 \theta = \frac{3}{5} \] Final Answer: \[ \sin^4 \theta - \cos^4 \theta = \frac{3}{5} \]