\( \lfloor \sum _ { i = 0 } ^ { 10 } \frac { i } { \cos \frac { 1 } { 10 } - \frac { \pi 1 } { 100 } } \rfloor - \lceil \int _ { - 2 \pi } ^ { 2 \pi } \frac { 3 \pi } { 10 } d x \rceil \frac { 1 } { \lfloor \sqrt { \ln ( 100 \operatorname { coth } \frac { \pi } { 8 } ) } \rfloor - ( \sin ^ { 2 } ( x ) + \cos ^ { 2 } ( x ) ) } \)

Let’s break down the expression into manageable parts to solve it. 1. **Sum Calculation**: The first part involves the sum \( \sum_{i=0}^{10} i \), which equals \( 0 + 1 + 2 + ... + 10 = \frac{10 \times 11}{2} = 55 \). Thus, we need to divide this sum by \( \cos \frac{1}{10} - \frac{\pi}{100} \). You would find \( \cos \frac{1}{10} \) and approximate to get a numerical value. 2. **Integral Calculation**: The second part involves the integral \( \int_{-2\pi}^{2\pi} \frac{3\pi}{10} \, dx \). This evaluates to \( \frac{3\pi}{10} \times (2\pi - (-2\pi)) = \frac{3\pi}{10} \times 4\pi = \frac{12\pi^2}{10} = \frac{6\pi^2}{5} \). 3. **Ceiling Function**: The next step requires finding \( \lceil \frac{6\pi^2}{5} \rceil \), which you can compute numerically. 4. **Square Root and Logarithm**: For the term \( \lfloor \sqrt{\ln(100 \operatorname{coth} \frac{\pi}{8})} \rfloor \), you must examine both \( \ln(100) \) and \( \operatorname{coth} \frac{\pi}{8} \), and compute a numerical approximation. 5. **Trigonometric Identity**: The term \( \sin^2(x) + \cos^2(x) \) always equals 1 for all \( x \). So you would revise each element, combining it all together at the end, and applying the floor and ceiling appropriately, to resolve the overall expression!