Trigonometric Equations Question 373
Question: The solution of the equation $ \sec \theta -cosec\theta =\frac{4}{3} $ is
[Roorkee 1994]
Options:
A) $ \frac{1}{2}[n\pi +{{(-1)}^{n}}{{\sin }^{-1}}(3/4)] $
B) $ n\pi +{{(-1)}^{n}}{{\sin }^{-1}}(3/4) $
C) $ \frac{n\pi }{2}+{{(-1)}^{n}}{{\sin }^{-1}}(3/4) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- $ 3(\sin \theta -\cos \theta )=4\sin \theta \cos \theta $
Þ $ 3(\sin \theta -\cos \theta )=2\sin 2\theta $ Squaring both sides, we get $ 9(1-S)=4S^{2}, $ where $ S=\sin 2\theta $ or $ 4S^{2}+9S-9=0 $ .
$ \therefore $ $ ,(S+3),(4S-3)=0 $ or $ S=\frac{3}{4} $ as $ S\ne -3 $ or $ \sin 2\theta =\frac{3}{4}=\sin \alpha $
$ \therefore $ $ 2\theta =n\pi +{{(-1)}^{n}}\alpha $ or $ \theta =\frac{1}{2},[ n\pi +{{(-1)}^{n}}{{\sin }^{-1}}( \frac{3}{4} ) ] $ .
BETA
