A) \[\frac{\pi }{10}\]
B) \[\frac{\pi }{5}\]
C) \[\frac{2\pi }{5}\]
D) \[\frac{4\pi }{5}\]
Correct Answer: C
Solution :
[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 \[{{\cos }^{-1}}\left( \frac{\sqrt{5}-1}{4} \right)={{\cos }^{-1}}\left( \sin \frac{\pi }{10} \right)\] \[={{\cos }^{-1}}\left( \cos \frac{2\pi }{5} \right)=\frac{2\pi }{5}\]
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