Question

The value of the expression $ 1-\frac{\sin^2y}{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)} $
$= 1-1+\cos y+0$
$= \cos y$