Find the values of cos−1x in terms of given options.
A
2sin−1√1−x2
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B
2cos−1√1−x2
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C
2cos−1√1+x2
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D
2sin−1√1+x2
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Solution
The correct options are A2sin−1√1−x2 C2cos−1√1+x2 Let cos−1x=y, so that x=cosy. Then ⇒√1−x2=√1−cosy2=√2sin2(y2)2=siny2⇒sin−1√1−x2=y2⇒2sin−1√1−x2=y Also √1+x2=√1+cosy2=√2cos2(y2)2=cosy2⇒cos−1√1+x2=y2⇒2cos−1√1+x2=y∴y=cos−1x=2sin−1√1−x2=2cos−1√1+x2