Trigonometric Identities Question 277
Question: General solution of the equation $ 2{{\cot }^{2}}\theta +2\sqrt{3}\cot \theta +4cosec+8=0 $ is
Options:
A) $ \theta =n\pi \pm \frac{\pi }{6},n\in I $
B) $ n\pi +\frac{\pi }{6},n\in I $
C) $ 2n\pi +\frac{\pi }{6},n\in I $
D) $ 2n\pi +\frac{11\pi }{6},n\in I $
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Answer:
Correct Answer: D
Solution:
$ 2{{\cot }^{2}}\theta +2\sqrt{3}\cot \theta +4\cos ec\theta +8=0 $
$ \Rightarrow {{\cot }^{2}}\theta +\cos ec^{2}\theta -1+2\sqrt{3}\cot \theta +4\cos ec\theta +8=0 $ $ ( \because ,{{\cot }^{2}}\theta =\cos ec^{2}\theta -1 ) $
$ \Rightarrow ( {{\cot }^{2}}\theta +2\sqrt{3}\cot \theta +3 )+( \cos ec^{2}\theta +4\cos ec\theta +4 )=0 $
$ \Rightarrow {{( \cot \theta +\sqrt{3} )}^{2}}+{{( \cos ec\theta +2 )}^{2}}=0 $
$ \Rightarrow ,\cot \theta =-\sqrt{3} $ and $ \cos ec\theta =-2 $ Principal value of $ \theta $ satisfying both the equation is $ \theta =2\pi -\frac{\pi }{6}=\frac{11\pi }{6} $
$ \therefore $ General solution is $ \theta =2n\pi +\frac{11\pi }{6},n\in I $
BETA
