1. **State the problem:** Prove that $$\frac{\cos 40^\circ - \sin 30^\circ}{\sin 60^\circ - \cos 50^\circ} = \tan 50^\circ$$. 2. **Recall important values and identities:** - $\sin 30^\circ = \frac{1}{2}$ - $\sin 60^\circ = \frac{\sqrt{3}}{2}$ - Use complementary angle identities: $\cos 40^\circ = \sin 50^\circ$ and $\cos 50^\circ = \sin 40^\circ$ 3. **Rewrite numerator and denominator using these identities:** $$\cos 40^\circ - \sin 30^\circ = \sin 50^\circ - \frac{1}{2}$$ $$\sin 60^\circ - \cos 50^\circ = \frac{\sqrt{3}}{2} - \sin 40^\circ$$ 4. **Substitute back:** $$\frac{\sin 50^\circ - \frac{1}{2}}{\frac{\sqrt{3}}{2} - \sin 40^\circ}$$ 5. **Express $\sin 40^\circ$ as $\sin(90^\circ - 50^\circ) = \cos 50^\circ$ and simplify denominator:** $$\frac{\sin 50^\circ - \frac{1}{2}}{\frac{\sqrt{3}}{2} - \cos 50^\circ}$$ 6. **Use exact values for $\cos 50^\circ$ and $\sin 50^\circ$ or use the sine and cosine subtraction formulas to simplify further. Alternatively, use the sine subtraction formula for numerator and denominator:** Numerator: $\sin 50^\circ - \sin 30^\circ = 2 \cos \frac{50^\circ + 30^\circ}{2} \sin \frac{50^\circ - 30^\circ}{2} = 2 \cos 40^\circ \sin 10^\circ$ Denominator: $\sin 60^\circ - \cos 50^\circ = \sin 60^\circ - \sin 40^\circ = 2 \cos \frac{60^\circ + 40^\circ}{2} \sin \frac{60^\circ - 40^\circ}{2} = 2 \cos 50^\circ \sin 10^\circ$ 7. **Simplify the fraction:** $$\frac{2 \cos 40^\circ \sin 10^\circ}{2 \cos 50^\circ \sin 10^\circ} = \frac{\cos 40^\circ}{\cos 50^\circ}$$ 8. **Recall that $\cos 40^\circ = \sin 50^\circ$, so:** $$\frac{\cos 40^\circ}{\cos 50^\circ} = \frac{\sin 50^\circ}{\cos 50^\circ} = \tan 50^\circ$$ **Final answer:** $$\frac{\cos 40^\circ - \sin 30^\circ}{\sin 60^\circ - \cos 50^\circ} = \tan 50^\circ$$