answer:Given the expression {eq}dfrac{1}{1+cos(x)}+dfrac{cos(x)}{1-cos(x)} {/eq}, we can simplify it as follows: {eq}begin{align*} left( dfrac{1}{1+cos(x)}+dfrac{cos(x)}{1-cos(x)} right) &= dfrac{1(1-cos(x)) + cos(x)(1+cos(x))}{(1+cos(x))(1-cos(x))} &= dfrac{1 - cos(x) + cos(x) + cos^2(x)}{1 - cos^2(x)} &= dfrac{1 + cos^2(x)}{sin^2(x)}quad text{(Using the identity} sin^2(x) + cos^2(x) = 1) &= dfrac{2 - sin^2(x)}{sin^2(x)} &= dfrac{2}{sin^2(x)} - dfrac{sin^2(x)}{sin^2(x)} &= 2csc^2(x) - 1 end{align*} {/eq} Hence, the expression simplified in terms of the cosecant function, {eq}csc(x) {/eq}, is {eq}2csc^2(x) - 1 {/eq}.

answer:The curl of the vector field vec{F} is given by the cross product of the partial derivatives of its components with respect to the other coordinates. The curl is calculated as follows: nabla times vec{F} = left( frac{partial h}{partial y} - frac{partial g}{partial z} right)uvec{i} - left( frac{partial f}{partial z} - frac{partial h}{partial x} right)uvec{j} + left( frac{partial g}{partial x} - frac{partial f}{partial y} right)uvec{k} Using the given functions: 1. frac{partial h}{partial y} = frac{3}{2}y^{1/2}, 2. frac{partial g}{partial z} = 0, 3. frac{partial f}{partial z} = 0, 4. frac{partial h}{partial x} = 0, 5. frac{partial g}{partial x} = 0, 6. frac{partial f}{partial y} = 1. Substituting these values, we get: nabla times vec{F} = left( frac{3}{2}y^{1/2} - 0 right)uvec{i} - (0 - 0)uvec{j} + (0 - 1)uvec{k} Simplifying, we obtain the curl of the vector field: nabla times vec{F} = left{ frac{3}{2}y^{1/2}, 0, -1 right}

answer:A gyrator is a conceptual electronic component that can emulate the behavior of a high-value inductor or capacitor. However, it does not physically store energy like an actual inductor or capacitor would. When simulating a 10H inductor using a gyrator, the energy storage aspect is not replicated, as the gyrator doesn't possess the capability to store energy in an electric or magnetic field. Instead, the gyrator provides a means to approximate the frequency response and quality factor (Q) of an inductor, but it does not offer the energy storage properties inherent to a genuine inductor.

answer:To find the number of pages Allison can fill, we need to divide the total number of photos by the number of photos that can fit on each page. Number of pages = Total number of photos / Number of photos per page Number of pages = 112 photos / 4 photos per page Number of pages = 28 pages Therefore, Allison can fill 28 pages of the photo album. Allison can fill 28 pages.