Q. 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) $\cos y$
Explanation
(d)
The given expression can be written as $\frac{1 + \cos{y - \sin^{2}y}}{1 + \cos y} + \frac{\left( 1 - \cos^{2}y \right) - \sin^{2}y}{\sin{y(1 - \cos y)}}$
$= \ \frac{\cos{y\ \left( 1 + \cos y \right)}}{1 + \cos y} + 0 = \cos y$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.