Quantitative Aptitude Ques 2288
Question: If $ \sin \theta +\sqrt{\sin \theta +\sqrt{\sin \theta +\sqrt{\sin \theta +….\infty }}}={{\sec }^{4}}\alpha , $ then $ \sin \theta $ is equal to
Options:
A) $ {{\sec }^{2}}\alpha $
B) $ {{\tan }^{2}}\alpha $
C) $ sec^{2}\alpha {{\tan }^{2}}\alpha $
D) $ {{\cos }^{2}}\alpha $
Show Answer
Answer:
Correct Answer: C
Solution:
- Given, $ \sin \theta +\sqrt{\sin \theta +\sqrt{\sin \theta +\sqrt{\sin \theta +…..\infty }}}={{\sec }^{4}}\alpha $ Then, $ \sin \theta +\sqrt{{{\sec }^{4}}\alpha }={{\sec }^{4}}\alpha $
$ \Rightarrow $ $ \sin \theta +{{\sec }^{2}}\alpha ={{\sec }^{4}}\alpha $
$ \Rightarrow $ $ \sin \theta ={{\sec }^{4}}\alpha -{{\sec }^{2}}\alpha $
$ \Rightarrow $ $ \sin \theta ={{\sec }^{2}}\alpha ,({{\sec }^{2}}\alpha -1) $
$ \Rightarrow $ $ \sin \theta ={{\sec }^{2}}\alpha \cdot {{\tan }^{2}}\alpha $ $ [\because {{\sec }^{2}}\alpha -1={{\tan }^{2}}\alpha ] $
BETA
