Trigonometric Identities Question 163
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
Options:
A) 0
B) 1
C) $ \sin ,y $
D) $ \cos ,y $
Show Answer
Answer:
Correct Answer: D
Solution:
The expression can be written as $ \frac{1+\cos y-{{\sin }^{2}}y}{1+\cos y}+\frac{(1-{{\cos }^{2}}y)-{{\sin }^{2}}y}{\sin y,(1-\cos y)} $ $ =\frac{\cos y,(1+\cos y)}{1+\cos y}+0=\cos y. $
BETA
