1. **State the problem:** Find equivalent ratios in terms of acute angles and then find the exact value of the expression $\tan 150^\circ - \sin 135^\circ$.\n\n2. **Recall angle relationships:**\n- $150^\circ = 180^\circ - 30^\circ$, so $\tan 150^\circ = \tan(180^\circ - 30^\circ)$.\n- $135^\circ = 180^\circ - 45^\circ$, so $\sin 135^\circ = \sin(180^\circ - 45^\circ)$.\n\n3. **Use trigonometric identities:**\n- $\tan(180^\circ - \theta) = -\tan \theta$.\n- $\sin(180^\circ - \theta) = \sin \theta$.\n\n4. **Apply these to the given angles:**\n$$\tan 150^\circ = -\tan 30^\circ$$\n$$\sin 135^\circ = \sin 45^\circ$$\n\n5. **Recall exact values for acute angles:**\n- $\tan 30^\circ = \frac{1}{\sqrt{3}}$.\n- $\sin 45^\circ = \frac{\sqrt{2}}{2}$.\n\n6. **Substitute these values:**\n$$\tan 150^\circ - \sin 135^\circ = -\frac{1}{\sqrt{3}} - \frac{\sqrt{2}}{2}$$\n\n7. **Simplify the expression:**\nTo combine, find a common denominator $2\sqrt{3}$:\n$$-\frac{1}{\sqrt{3}} = -\frac{2}{2\sqrt{3}}$$\n$$-\frac{\sqrt{2}}{2} = -\frac{\sqrt{2}\sqrt{3}}{2\sqrt{3}} = -\frac{\sqrt{6}}{2\sqrt{3}}$$\n\nSo,\n$$-\frac{1}{\sqrt{3}} - \frac{\sqrt{2}}{2} = -\frac{2}{2\sqrt{3}} - \frac{\sqrt{6}}{2\sqrt{3}} = -\frac{2 + \sqrt{6}}{2\sqrt{3}}$$\n\n8. **Rationalize the denominator:**\nMultiply numerator and denominator by $\sqrt{3}$:\n$$-\frac{2 + \sqrt{6}}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = -\frac{(2 + \sqrt{6})\sqrt{3}}{2 \times 3} = -\frac{(2 + \sqrt{6})\sqrt{3}}{6}$$\n\n**Final exact value:**\n$$\boxed{-\frac{(2 + \sqrt{6})\sqrt{3}}{6}}$$