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

Answer 1

The correct option is D $\cos y$

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

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

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

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

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

$\Rightarrow \cos y\frac{(1+\cos y)}{(1+\cos y)}=\cos y$.