Trigonometric Identities Question 147
Question: If $ \theta $ lies in the second quadrant, then the value of $ \sqrt{( \frac{1-\sin \theta }{1+\sin \theta } )}+\sqrt{( \frac{1+\sin \theta }{1-\sin \theta } )} $
Options:
A) $ 2\sec \theta $
B) $ -2\sec \theta $
C) $ 2cosec\theta $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \sqrt{( \frac{1-\sin \theta }{1+\sin \theta } )}+\sqrt{( \frac{1+\sin \theta }{1-\sin \theta } )} $ is the sum of two positive quantities and hence the result must be positive. But for $ \frac{\pi }{2}<\theta <\pi , $ we have the sum equal to $ \frac{1-\sin \theta +1+\sin \theta }{\sqrt{1-{{\sin }^{2}}\theta }}=\frac{2}{\cos \theta }; $ which is negative. ( $ \because $ $ \cos \theta $ is negative for q lying in 2nd quadrant). So the required positive value $ =\frac{-2}{\cos \theta }=-2,\sec \theta ,,( \frac{\pi }{2}<\theta <\pi ) $ .
BETA
