A) \[-\frac{1}{2\sqrt{1-{{x}^{2}}}}\]
B) \[\frac{1}{2\sqrt{1-{{x}^{2}}}}\]
C) \[\frac{1}{\sqrt{1-x}}\]
D) \[{{\sin }^{-1}}\left\{ \sqrt{\frac{1+x}{2}} \right\}\]
Correct Answer: A
Solution :
\[y={{\cos }^{-1}}\left\{ \sqrt{\frac{1+x}{2}} \right\}\] Let \[\sqrt{\frac{1+x}{2}}=\cos \theta \] or \[x=2{{\cos }^{2}}\theta -1=\cos 2\theta \]; \[\therefore \theta =\frac{1}{2}{{\cos }^{-1}}x\]. So, \[y=\frac{1}{2}{{\cos }^{-1}}x\]Þ \[-\frac{1}{2\sqrt{1-{{x}^{2}}}}\].
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