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Question: The value of the expression \(1 - \frac{\sin^{2}y}{1 + \cos y} + \frac{1 + \cos y}{\sin y} - \frac{...

The value of the expression

1−sin⁡2y1+cos⁡y+1+cos⁡ysin⁡y−sin⁡y1−cos⁡y1 - \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\sin y

D

cos⁡y\cos y

Answer

cos⁡y\cos y

Explanation

Solution

The expression can be written as

1+cos⁡y−sin⁡2y1+cos⁡y+(1−cos⁡2y)−sin⁡2ysin⁡y(1−cos⁡y)\frac{1 + \cos y - \sin^{2}y}{1 + \cos y} + \frac{(1 - \cos^{2}y) - \sin^{2}y}{\sin y(1 - \cos y)}

=cos⁡y(1+cos⁡y)1+cos⁡y+0=cos⁡y.= \frac{\cos y(1 + \cos y)}{1 + \cos y} + 0 = \cos y.