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 ] $

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