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B
0
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C
1
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D
−1
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Solution
The correct option is C−12 Tofindddx(tan−1√1+cosx1−cosx)ByApplyingchainruleletustakeu=√1+cosx1−cosxThereforeddu(tan−1u)⋅ddx(√1+cosx1−cosx)Step1:ddutan−1u=1u2+1Step2:ddx√1+cosx1−cosx=−sinx√1+cosx⋅(1−cosx)3/2CombiningStep1andStep2=1u2+1⋅(−sinx√1+cosx⋅(1−cosx)3/2)substitutingthevalueof′u′inthegivenequation=1(√1+cosx1−cosx)2+1⋅(−sinx√1+cosx⋅(1−cosx)3/2)=−sinx2√1+cosx(1−cosx)1/2Since0<x<πLetustakethevalueasπ2Sothevaluewouldbe=−sinπ22√1+cosπ2(1−cosπ2)12=−12√1+0(1−0)1/2=−12Thereforeddx(tan−1√1+cosx1−cosx)=−12at0<x<π