1. **Stating the problem:** Simplify the expression $$\sin(90^\circ - \alpha) + \tan(54^\circ - \alpha) - \cos(90^\circ - \alpha) + \sin(-\alpha) + \sin(\alpha - 120^\circ) - \tan(-\alpha)$$ 2. **Recall trigonometric identities:** - $\sin(90^\circ - x) = \cos x$ - $\cos(90^\circ - x) = \sin x$ - $\sin(-x) = -\sin x$ - $\tan(-x) = -\tan x$ - $\tan(180^\circ - x) = -\tan x$ - $\sin(180^\circ - x) = \sin x$ 3. **Apply identities to each term:** $$\sin(90^\circ - \alpha) = \cos \alpha$$ $$- \cos(90^\circ - \alpha) = -\sin \alpha$$ $$\sin(-\alpha) = -\sin \alpha$$ $$- \tan(-\alpha) = + \tan \alpha$$ So the expression becomes: $$\cos \alpha + \tan(54^\circ - \alpha) - \sin \alpha - \sin \alpha + \sin(\alpha - 120^\circ) + \tan \alpha$$ 4. **Group like terms:** $$\cos \alpha - 2\sin \alpha + \tan(54^\circ - \alpha) + \sin(\alpha - 120^\circ) + \tan \alpha$$ 5. **Note:** Without specific values for $\alpha$, this is the simplified form using identities. 6. **If further simplification is needed, use angle subtraction formulas:** $$\sin(\alpha - 120^\circ) = \sin \alpha \cos 120^\circ - \cos \alpha \sin 120^\circ = \sin \alpha \times (-\frac{1}{2}) - \cos \alpha \times \frac{\sqrt{3}}{2} = -\frac{1}{2} \sin \alpha - \frac{\sqrt{3}}{2} \cos \alpha$$ 7. **Substitute back:** $$\cos \alpha - 2\sin \alpha + \tan(54^\circ - \alpha) - \frac{1}{2} \sin \alpha - \frac{\sqrt{3}}{2} \cos \alpha + \tan \alpha$$ 8. **Combine like terms:** $$\left(\cos \alpha - \frac{\sqrt{3}}{2} \cos \alpha\right) + \left(-2\sin \alpha - \frac{1}{2} \sin \alpha\right) + \tan(54^\circ - \alpha) + \tan \alpha$$ $$= \left(1 - \frac{\sqrt{3}}{2}\right) \cos \alpha - \frac{5}{2} \sin \alpha + \tan(54^\circ - \alpha) + \tan \alpha$$ This is the simplified expression. **Final answer:** $$\boxed{\left(1 - \frac{\sqrt{3}}{2}\right) \cos \alpha - \frac{5}{2} \sin \alpha + \tan(54^\circ - \alpha) + \tan \alpha}$$