Let α,β(α>β) be the roots of the equation sin−1x−1sin−1x=cos−1x−1cos−1x. Then
A
α=1√2
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B
β=−1√2
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C
sin−1α=π3
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D
sin−1β=π4−√1+π216
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Solution
The correct options are Aα=1√2 Dsin−1β=π4−√1+π216 sin−1x−1sin−1x=cos−1x−1cos−1x⇒sin−1x−cos−1x+1cos−1x−1sin−1x=0⇒(sin−1x−cos−1x)(1+1cos−1xsin−1x)=0⇒(sin−1x−cos−1x)(sin−1xcos−1x+1)(sin−1x)(cos−1x)=0⇒sin−1x=cos−1x ⇒sin−1x=π2−sin−1x⇒sin−1x=π4⇒x=1√2 or sin−1xcos−1x+1=0⇒sin−1x(π2−sin−1x)+1=0⇒(sin−1x)2−π2(sin−1x)−1=0⇒sin−1x=π2±√(π24+4)2⇒sin−1x=π4−√(1+π216)