If cos(x−y),cosx,cos(x+y) are in H.P., where y≠2nπ,n∈Z, then the value of [cosxsecy2] is/are
(where [.] denotes greatest integer function)
A
−2
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B
−1
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C
0
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D
1
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Solution
The correct options are A−2 D1 cos(x−y),cosx,cos(x+y) are in H.P.
⇒cosx=2cos(x−y)cos(x+y)cos(x−y)+cos(x+y)⇒cosx=2(cos2x−sin2y)2cosxcosy⇒cos2xcosy=cos2x−sin2y⇒sin2y=cos2x(1−cosy)⇒1−cos2y=cos2x(1−cosy)⇒(1−cosy)(1+cosy)=cos2x(1−cosy)⇒1+cosy=cos2x(∵y≠2nπ⇒cosy≠1)⇒cos2x=2cos2y2⇒cos2xsec2y2=2⇒cosxsecy2=±√2⇒[cosx×secy2]=[√2] or [−√2]∴[cosx×secy2]=1,−2