Trigonometric Identities Question 311
Question: If $ \pi <\alpha <\frac{3\pi }{2} $ , then $ \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }}+\sqrt{\frac{1+\cos \alpha }{1-\cos \alpha }} $ =
Options:
A) $ \frac{2}{\sin \alpha } $
B) $ -\frac{2}{\sin \alpha } $
C) $ \frac{1}{\sin \alpha } $
D) $ -\frac{1}{\sin \alpha } $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }}+\sqrt{\frac{1+\cos \alpha }{1-\cos \alpha }}=\frac{1-\cos \alpha +1+\cos \alpha }{\sqrt{1-{{\cos }^{2}}\alpha }} $ $ =\frac{2}{\pm \sin \alpha } $ $ =\frac{2}{-\sin \alpha },( \text{since }\pi <\alpha <\frac{3\pi }{2} ). $
BETA
