Inverse Trigonometric Functions Question 87
Question: There exists a positive real number x satisfying $ \cos (ta{n^{-1}}x)=x $ , Then the value of $ {{\cos }^{-1}}( \frac{x^{2}}{2} ) $ is
Options:
A) $ \frac{\pi }{10} $
B) $ \frac{\pi }{5} $
C) $ \frac{2\pi }{5} $
D) $ \frac{4\pi }{5} $
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Answer:
Correct Answer: C
Let $ {{\tan }^{-1}}(x)=\theta $
Or $ x=\tan \theta $
$ \Rightarrow \cos \theta =x\Rightarrow \frac{1}{\sqrt{1+x^{2}}}=x $
Or $ x^{2}(1+x^{2})=1 $
Or $ x^{2}=\frac{-1\pm \sqrt{5}}{2} $
Or $ x^{2}=\frac{\sqrt{5}-1}{2} $ or $ \frac{x^{2}}{2}=\frac{\sqrt{5}-1}{4} $
Now $ {{\cos }^{-1}}( \frac{\sqrt{5}-1}{4} )={{\cos }^{-1}}( \sin \frac{\pi }{10} ) $
$ ={{\cos }^{-1}}( \cos \frac{2\pi }{5} )=\frac{2\pi }{5} $
BETA
