c. \( \cot ^{2} \frac{\pi^{c}}{6}-2 \cos ^{2} \frac{\pi^{c}}{3}-\frac{3}{4} \sec ^{2} \frac{\pi^{c}}{4}-4 \sec ^{2} \frac{\pi^{c}}{6} \)

To simplify the expression \[ \cot^2 \frac{\pi^c}{6} - 2 \cos^2 \frac{\pi^c}{3} - \frac{3}{4} \sec^2 \frac{\pi^c}{4} - 4 \sec^2 \frac{\pi^c}{6}, \] we will evaluate each term separately. 1. **Evaluate \( \cot^2 \frac{\pi^c}{6} \)**: \[ \cot \frac{\pi^c}{6} = \frac{\cos \frac{\pi^c}{6}}{\sin \frac{\pi^c}{6}}. \] Using the values: \[ \cos \frac{\pi^c}{6} = \frac{\sqrt{3}}{2}, \quad \sin \frac{\pi^c}{6} = \frac{1}{2}, \] we find: \[ \cot \frac{\pi^c}{6} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}. \] Therefore, \[ \cot^2 \frac{\pi^c}{6} = 3. \] 2. **Evaluate \( -2 \cos^2 \frac{\pi^c}{3} \)**: \[ \cos \frac{\pi^c}{3} = \frac{1}{2}, \] so: \[ \cos^2 \frac{\pi^c}{3} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}. \] Thus, \[ -2 \cos^2 \frac{\pi^c}{3} = -2 \cdot \frac{1}{4} = -\frac{1}{2}. \] 3. **Evaluate \( -\frac{3}{4} \sec^2 \frac{\pi^c}{4} \)**: \[ \sec \frac{\pi^c}{4} = \frac{1}{\cos \frac{\pi^c}{4}} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}, \] so: \[ \sec^2 \frac{\pi^c}{4} = 2. \] Therefore, \[ -\frac{3}{4} \sec^2 \frac{\pi^c}{4} = -\frac{3}{4} \cdot 2 = -\frac{3}{2}. \] 4. **Evaluate \( -4 \sec^2 \frac{\pi^c}{6} \)**: \[ \sec \frac{\pi^c}{6} = \frac{1}{\cos \frac{\pi^c}{6}} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}, \] so: \[ \sec^2 \frac{\pi^c}{6} = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}. \] Therefore, \[ -4 \sec^2 \frac{\pi^c}{6} = -4 \cdot \frac{4}{3} = -\frac{16}{3}. \] Now, we can combine all the evaluated terms: \[ 3 - \frac{1}{2} - \frac{3}{2} - \frac{16}{3}. \] First, combine \( 3 - \frac{1}{2} - \frac{3}{2} \): \[ 3 - 2 = 1. \] Now, we have: \[ 1 - \frac{16}{3} = \frac{3}{3} - \frac{16}{3} = -\frac{13}{3}. \] Thus, the final result is: \[ \boxed{-\frac{13}{3}}. \]