4. Simplify: 4.1 \( \frac{\operatorname{Sin} 120 \cdot \cos 210 \cdot \tan 515 \cdot \cos 27}{\sin 63 \cdot \cos 540} \)

We start with the given expression \[ \frac{\sin 120 \cdot \cos 210 \cdot \tan 515 \cdot \cos 27}{\sin 63 \cdot \cos 540}. \] **Step 1. Simplify \(\sin 120\):** Since \[ \sin 120 = \sin(180-60) = \sin 60 = \frac{\sqrt{3}}{2}, \] we have \[ \sin 120 = \frac{\sqrt{3}}{2}. \] **Step 2. Simplify \(\cos 210\):** Note that \[ 210 = 180 + 30, \] so \[ \cos 210 = -\cos 30 = -\frac{\sqrt{3}}{2}. \] **Step 3. Simplify \(\tan 515\):** Since the tangent function has period \(180^\circ\), \[ \tan 515 = \tan (515-360) = \tan 155. \] Express \(155\) as \(180-25\): \[ \tan 155 = \tan (180-25) = -\tan 25. \] Thus, \[ \tan 515 = -\tan 25. \] **Step 4. Note the remaining terms:** \(\cos 27\) remains as is. For the denominator, notice: \[ \sin 63 \quad \text{(remains)} \quad \text{and} \quad \cos 540. \] Since \[ 540-360=180, \] we have \[ \cos 540 = \cos 180 = -1. \] **Step 5. Substitute into the given expression:** Replacing each term, we obtain \[ \frac{\left(\frac{\sqrt{3}}{2}\right) \cdot \left(-\frac{\sqrt{3}}{2}\right) \cdot \left(-\tan 25\right) \cdot \cos 27}{\sin 63 \cdot (-1)}. \] **Step 6. Simplify the numerator:** First, multiply the sine and cosine factors: \[ \frac{\sqrt{3}}{2} \cdot \left(-\frac{\sqrt{3}}{2}\right) = -\frac{3}{4}. \] Then multiplying by \(-\tan 25\) yields: \[ -\frac{3}{4} \cdot (-\tan 25) = \frac{3}{4}\tan 25. \] Thus, the numerator becomes \[ \frac{3}{4}\tan 25 \cdot \cos 27. \] **Step 7. Simplify the denominator:** \(\sin 63 \cdot (-1) = -\sin 63.\) Now the expression is \[ \frac{\frac{3}{4}\tan 25 \cdot \cos 27}{-\sin 63}. \] **Step 8. Use the co-function identity:** Notice that \[ \sin 63 = \cos (90 - 63) = \cos 27. \] Thus, the expression simplifies to \[ \frac{\frac{3}{4}\tan 25 \cdot \cos 27}{- \cos 27} = -\frac{3}{4}\tan 25. \] The final simplified expression is \[ -\frac{3}{4}\tan 25. \]