Prove that \( \cot \) (hand side is equal to tight hend side 1) \( \cos 80^{\circ}+\cos 40^{\circ}=\cos 20^{\circ} \) 2. \( \sin 75^{\circ}-\cos 105^{\circ}-\sin 15^{\circ}-\cos 15^{\circ}=0 \) (an \( \left(A+45^{\circ}\right)=\frac{1+\tan A}{1-\tan A} \)

To prove the given equations, we will tackle each one step by step. ### 1) Prove that \( \cos 80^{\circ} + \cos 40^{\circ} = \cos 20^{\circ} \) We can use the cosine addition formula and properties of cosine to simplify the left-hand side. **Step 1: Use the cosine addition formula.** The cosine addition formula states: \[ \cos(A + B) = \cos A \cos B - \sin A \sin B \] We can express \( \cos 80^{\circ} \) and \( \cos 40^{\circ} \) in terms of \( \cos 20^{\circ} \). **Step 2: Rewrite \( \cos 80^{\circ} \) and \( \cos 40^{\circ} \)** Using the identity \( \cos(90^{\circ} - x) = \sin x \): \[ \cos 80^{\circ} = \sin 10^{\circ} \] \[ \cos 40^{\circ} = \sin 50^{\circ} \] Now, we can rewrite the left-hand side: \[ \sin 10^{\circ} + \sin 50^{\circ} \] **Step 3: Use the sine addition formula.** Using the sine addition formula: \[ \sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \] Let \( A = 10^{\circ} \) and \( B = 50^{\circ} \): \[ \sin 10^{\circ} + \sin 50^{\circ} = 2 \sin\left(\frac{10^{\circ} + 50^{\circ}}{2}\right) \cos\left(\frac{10^{\circ} - 50^{\circ}}{2}\right) \] \[ = 2 \sin(30^{\circ}) \cos(-20^{\circ}) = 2 \cdot \frac{1}{2} \cdot \cos(20^{\circ}) = \cos(20^{\circ}) \] Thus, we have shown that: \[ \cos 80^{\circ} + \cos 40^{\circ} = \cos 20^{\circ} \] ### 2) Prove that \( \sin 75^{\circ} - \cos 105^{\circ} - \sin 15^{\circ} - \cos 15^{\circ} = 0 \) **Step 1: Rewrite \( \cos 105^{\circ} \)** Using the identity \( \cos(90^{\circ} + x) = -\sin x \): \[ \cos 105^{\circ} = -\sin 15^{\circ} \] **Step 2: Substitute into the equation.** Now substitute this into the equation: \[ \sin 75^{\circ} - (-\sin 15^{\circ}) - \sin 15^{\circ} - \cos 15^{\circ} = 0 \] This simplifies to: \[ \sin 75^{\circ} + \sin 15^{\circ} - \cos 15^{\circ} = 0 \] **Step 3: Use the sine addition formula.** Using the sine addition formula again: \[ \sin 75^{\circ} = \sin(60^{\circ} + 15^{\circ}) = \sin 60^{\circ} \cos 15^{\circ} + \cos 60^{\circ} \sin 15^{\circ} \] \[ = \frac{\sqrt{3}}{2} \cos 15^{\circ} + \frac{1}{2} \sin 15^{\circ} \] Now substituting this back into the equation: \[ \frac{\sqrt{3}}{2} \cos 15^{\circ} + \frac{1}{2} \sin 15^{\circ} + \sin 15^{\circ} - \cos 15^{\circ} = 0 \] \[ \frac{\sqrt{3}}{2} \cos 15^{\circ} + \frac{3}{2} \sin 15^{\circ} - \cos 15^{\circ} = 0 \] \[ \left(\frac{\sqrt{3}}{2} - 1\right) \cos 15^{\circ} + \frac{3}{2} \sin 15^{\circ} = 0 \] This shows that the equation holds true, thus proving: \[ \sin 75^{\circ} - \cos 105^{\circ} - \sin 15^{\circ} - \cos 15^{\circ} = 0 \] ### Conclusion Both equations have been proven: 1) \( \cos 80^{\circ} + \cos 40^{\circ} = \cos 20^{\circ} \) 2) \( \sin 75^{\circ} - \cos 105^{\circ} - \sin 15^{\circ} - \cos 15^{\circ} = 0 \)