Question

$1-\frac{{\sin }^{2}y}{1+\cos y}+\frac{1+\cos y}{\sin y}-\frac{\sin y}{1-\cos y}=$

• A. 0
• B. 1
• C. $cosy$
• D. $siny$

Answer 1

The correct option is C $cosy$

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

Hence, the option $C$ is the correct answer.