A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{2}\]
Correct Answer: C
Solution :
Given that \[{{\cos }^{-1}}x=\alpha ,\] \[0<x<1\] ?.(i) \[\Rightarrow \] \[x=\cos \alpha \] \[\therefore \]\[{{\sin }^{-1}}(2x\sqrt{1-{{x}^{2}}})+{{\sec }^{-1}}\left( \frac{1}{2{{x}^{2}}-1} \right)=\frac{2\pi }{3}\] \[\Rightarrow \] \[{{\sin }^{-1}}(2\,\,\cos \alpha \sqrt{1-{{\cos }^{2}}\,\alpha )}\] \[+{{\sec }^{-1}}\left( \frac{1}{2{{\cos }^{2}}\,\alpha -1} \right)=\frac{2\pi }{3}\] \[\Rightarrow \] \[{{\sin }^{-1}}(\sin \,2\alpha )+{{\sec }^{-1}}(\sec \,2\alpha )=\frac{2\pi }{3}\] \[\Rightarrow \] \[2\alpha +2\alpha =\frac{2\pi }{3}\] \[\Rightarrow \] \[\alpha =\frac{\pi }{6}\] From (i) Now, \[x=\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}\] \[\Rightarrow \] \[2x=\sqrt{3}\] \[\therefore \] \[{{\tan }^{-1}}(2x)={{\tan }^{-1}}(\sqrt{3})\] \[=\frac{\pi }{3}\]
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