Quantitative Aptitude Ques 686
Question: If $ \sec \theta +\tan \theta =2+\sqrt{3}, $ then the value of $ \sec \theta $ is
Options:
A) $ \sqrt{3} $
B) $ 2 $
C) $ 4 $
D) $ 2\sqrt{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \sec \theta +\tan \theta =2+\sqrt{3} $ On squaring both sides, we get $ {{\sec }^{2}}\theta +{{\tan }^{2}}\theta +2\sec \theta \tan \theta =4+3+4\sqrt{3} $
$ \Rightarrow $ $ {{\sec }^{2}}\theta +{{\sec }^{2}}\theta -1+2\sec \theta \tan \theta =7+4\sqrt{3} $
$ \Rightarrow $ $ 2{{\sec }^{2}}\theta +2\sec \theta \tan \theta =8+4\sqrt{3} $
$ \Rightarrow $ $ 2\sec \theta (\sec \theta +\tan \theta )=4(2+\sqrt{3)} $
$ \therefore $ $ \sec \theta =\frac{2(2+\sqrt{3})}{\sec \theta +\tan \theta }=\frac{2(2+\sqrt{3})}{(2+\sqrt{3})}=2 $
BETA
