Question

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

• A. $\cos y$
• B. 1
• C. 0
• D. $\sin y$

Answer 1

Answer: (A)

$1-\frac{{\sin }^{2}y}{1+\cos y}+\frac{1+\cos y}{\sin y}-\frac{\sin y}{1-\cos y}$
$=1-\left(\frac{1-{\cos }^{2}y}{1+\cos y}\right)+\frac{1-{\cos }^{2}y-{\sin }^{2}y}{\sin y\left(1-\cos y\right)}$