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

Answer 1

Answer: (B)

We have,
$1-\frac{{\sin }^{2}y}{1+\cos y}+\frac{1+\cos y}{\sin y}-\frac{\sin y}{1-\cos y}$
$=1-\frac{\left(1-{\cos }^{2}y\right)}{1+\cos y}+\frac{(1+\cos y)(1-\cos y)-{\sin }^{2}y}{\sin y(1-\cos y)}$
$=1-\frac{(1-\cos y)(1+\cos y)}{(1+\cos y)}+\frac{\left(1-{\cos }^{2}y\right)-{\sin }^{2}y}{\sin y(1-\cos y)}$
$\left[\because a^2-b^2=(a-b)(a+b)\right]$
$=1-(1-\cos y)+\frac{{\sin }^{2}y-{\sin }^{2}y}{\sin y(1-\cos y)}$
$=1-1+\cos y+0$
$=\cos y$