SATHEE: Quantitative Aptitude Ques 1195

Quantitative Aptitude Ques 1195

Question: If $ \sin \theta =\frac{a^{2}-1}{a^{2}+1}, $ then the value of $ \sec \theta +\tan \theta $ will be

Options:

A) $ \frac{a}{\sqrt{2}} $

B) $ \frac{a}{a^{2}+1} $

C) $ \sqrt{2}a $

D) $ a $

Show Answer

Answer:

Correct Answer: D

Solution:

$ \sin \theta =\frac{a^{2}-1}{a^{2}+1} $ In $ \Delta ABC, $ $ BC=\sqrt{{{(AC)}^{2}}-{{(AB)}^{2}}} $ $ =\sqrt{{{(a^{2}+1)}^{2}}-{{(a^{2}-1)}^{2}}} $ $ =\sqrt{a^{4}+1+2a^{2}-a^{4}-1+2a^{2}} $ $ =\sqrt{4a^{2}}=2a $

$ \therefore $ $ \sec \theta +tan\theta =\frac{a^{2}+1}{2a}+\frac{a^{2}-1}{2a} $ $ =\frac{a^{2}+1+a^{2}-1}{2a}=\frac{2a^{2}}{2a}=a $

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Quantitative Aptitude Ques 1195

Question: If $ \sin \theta =\frac{a^{2}-1}{a^{2}+1}, $ then the value of $ \sec \theta +\tan \theta $ will be

Options:

Answer:

Solution:

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