The value of the expression $1-\frac{{\sin }^2\mathrm{y}}{1+\cos \mathrm{y}}+\frac{1+\cos \mathrm{y}}{\sin \mathrm{y}}-\frac{\sin \mathrm{y}}{1-\cos \mathrm{y}}$ is equal to

Solution:

The given expression can be written as

$\begin{array}{l}\frac{1+\cos \mathrm{y}-{\sin }^2\mathrm{y}}{1+\cos \mathrm{y}}+\frac{\left(1-{\cos }^2\mathrm{y}\right)-{\sin }^2\mathrm{y}}{\sin \mathrm{y}(1-\cos \mathrm{y})}\\ =\frac{\cos \mathrm{y}(1+\cos \mathrm{y})}{1+\cos \mathrm{y}}+0=\cos \mathrm{y}\end{array}$