The value of the expression 1−sin2⁡y1+cos⁡y+1+cos⁡ysin⁡y−sin⁡y1−cos⁡y is equal to

# The value of the expression $1-\frac{{\mathrm{sin}}^{2}\mathrm{y}}{1+\mathrm{cos}\mathrm{y}}+\frac{1+\mathrm{cos}\mathrm{y}}{\mathrm{sin}\mathrm{y}}-\frac{\mathrm{sin}\mathrm{y}}{1-\mathrm{cos}\mathrm{y}}$ is equal to

1. A

0

2. B

1

3. C

sin y

4. D

cos y

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### Solution:

The given expression can be written as

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

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