1. **State the problem:** Show that $$\tan\left(\frac{45^\circ + \theta}{2}\right) \cdot \tan\left(\frac{45^\circ - \theta}{2}\right) = \frac{\sqrt{2} \cos \theta - 1}{\sqrt{2} \cos \theta + 1}$$. 2. **Use tangent sum and difference formulas for half angles:** Let $$A = \frac{45^\circ + \theta}{2}$$ and $$B = \frac{45^\circ - \theta}{2}$$. Then the product is $$\tan A \cdot \tan B$$. 3. **Express tangent in terms of sine and cosine:** $$\tan A \cdot \tan B = \frac{\sin A}{\cos A} \cdot \frac{\sin B}{\cos B} = \frac{\sin A \sin B}{\cos A \cos B}.$$ 4. **Use sine and cosine addition/subtraction formulas:** Recall that $$A + B = 45^\circ$$, so $$\sin A \sin B = \frac{\cos(A - B) - \cos(A + B)}{2}$$ and $$\cos A \cos B = \frac{\cos(A - B) + \cos(A + B)}{2}.$$ 5. **Substitute $A \pm B$ values:** Since $$A - B = \frac{(45^\circ + \theta) - (45^\circ - \theta)}{2} = \frac{2\theta}{2} = \theta$$ and $$A + B = 45^\circ$$ we have: $$\sin A \sin B = \frac{\cos \theta - \cos 45^\circ}{2}, \quad \cos A \cos B = \frac{\cos \theta + \cos 45^\circ}{2}.$$ 6. **Form the ratio:** $$\tan A \tan B = \frac{\sin A \sin B}{\cos A \cos B} = \frac{\cos \theta - \cos 45^\circ}{\cos \theta + \cos 45^\circ}.$$ 7. **Substitute $$\cos 45^\circ = \frac{\sqrt{2}}{2}$$:** $$\tan A \tan B = \frac{\cos \theta - \frac{\sqrt{2}}{2}}{\cos \theta + \frac{\sqrt{2}}{2}}.$$ 8. **Multiply numerator and denominator by $$\sqrt{2}$$ to clear denominators:** $$\tan A \tan B = \frac{\sqrt{2} \cos \theta - 1}{\sqrt{2} \cos \theta + 1}.$$ **Final answer:** $$\tan\left(\frac{45^\circ + \theta}{2}\right) \cdot \tan\left(\frac{45^\circ - \theta}{2}\right) = \frac{\sqrt{2} \cos \theta - 1}{\sqrt{2} \cos \theta + 1}.$$