Trigonometric Equations Question 31
Question: The number of solutions of the given equation $ \tan \theta +\sec \theta =\sqrt{3}, $ where $ 0<\theta <2\pi $ is
Options:
A) 0
B) 1
C) 2
D) 3
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \sec \theta +\tan \theta =\sqrt{3} $ ?.(i) Also we have $ {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 $ ?..(ii)
$ \Rightarrow $ $ \sec \theta -\tan \theta =\frac{1}{\sqrt{3}} $ ?..(iii) Now (i) and (iii) gives, $ \tan \theta =\frac{1}{2}( \sqrt{3}-\frac{1}{\sqrt{3}} )=\frac{1}{\sqrt{3}}=\tan ( \frac{\pi }{6} ) $
$ \Rightarrow $ $ \theta =n\pi +\frac{\pi }{6} $ .
$ \therefore $ Solutions for $ 0\le \theta \le 2\pi $ are $ \frac{\pi }{6} $ and $ \frac{7\pi }{6} $ . Hence there are two solutions.
BETA
