A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{6}\]
Correct Answer: D
Solution :
Consider\[{{\tan }^{-1}}\left[ \sin \left( {{\cos }^{-1}}\sqrt{\frac{2}{3}} \right) \right]\] Let\[{{\cos }^{-1}}\sqrt{\frac{2}{3}}=\theta \Rightarrow \cos \theta =\sqrt{\frac{2}{3}}\] \[\Rightarrow \]\[\sin \theta \sqrt{1-{{\cos }^{2}}\theta }=\sqrt{1-\frac{2}{3}}=\sqrt{\frac{1}{3}}\] \[\therefore \]\[\left[ \sin \left( {{\cos }^{-1}}\sqrt{\frac{2}{3}} \right) \right]={{\tan }^{-1}}[\sin \theta ]\] \[=\left[ \sqrt{\frac{1}{3}} \right]={{\tan }^{-1}}\left( \frac{1}{\sqrt{3}} \right)\]\[=\frac{\pi }{6}\]
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