6. Simplify: \( \frac{\cos 68^{\circ}}{\cos 34^{\circ}-\sin 34^{\circ}} \)

To simplify the expression \( \frac{\cos 68^{\circ}}{\cos 34^{\circ}-\sin 34^{\circ}} \), we first recognize that \( \cos 68^{\circ} = \sin 22^{\circ} \) (since \( \cos(90^{\circ} - x) = \sin x \)). Next, we simplify the denominator: \[ \cos 34^{\circ} - \sin 34^{\circ} = \cos 34^{\circ} - \cos(90^{\circ} - 34^{\circ}) = \cos 34^{\circ} - \sin 34^{\circ}. \] Now, we can factor the denominator using the identity \( a - b = \sqrt{2} \sin\left(\frac{\pi}{4} - x\right) \). Thus, we utilize the fact that: \[ \cos 34^{\circ} - \sin 34^{\circ} = \sqrt{2} \left( \cos 34^{\circ} \cdot \frac{1}{\sqrt{2}} - \sin 34^{\circ} \cdot \frac{1}{\sqrt{2}} \right) = \sqrt{2} \sin\left( 45^{\circ} - 34^{\circ} \right) = \sqrt{2} \sin(11^{\circ}). \] Let's rewrite the whole expression: \[ \frac{\cos 68^{\circ}}{\cos 34^{\circ} - \sin 34^{\circ}} = \frac{\sin 22^{\circ}}{\sqrt{2} \sin 11^{\circ}}. \] Using the identity \( \sin 22^{\circ} = 2 \sin 11^{\circ} \cos 11^{\circ} \), we substitute this in: \[ \frac{2 \sin 11^{\circ} \cos 11^{\circ}}{\sqrt{2} \sin 11^{\circ}}. \] Assuming \( \sin 11^{\circ} \neq 0 \), we can simplify by cancelling \( \sin 11^{\circ} \) in the numerator and denominator: \[ = \frac{2 \cos 11^{\circ}}{\sqrt{2}} = \frac{2 \cos 11^{\circ}}{\sqrt{2}} = \sqrt{2} \cos 11^{\circ}. \] Thus, the simplified expression is: \[ \sqrt{2} \cos 11^{\circ}. \]