A) \[\pi /6\]
B) \[\pi /4\]
C) \[\pi /3\]
D) \[\pi /2\]
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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