Column IColumn 2Column 3(I)cos(sin−1(sin5π6))=(i)1(P)f(x)=3x3−13x2+14x−2 hasthree roots α,β,γ, then value of|sin(tan−1α+tan−1β+tan−1γ)|=(II)cos(cos−1(−sin7π6))=(ii)1√2(Q)(√3cosec20∘−sec20∘)8=(III)∣∣cos{tan−1(tan15π4)}∣∣=(iii)12(R)4√2(sin12∘)(sin48∘)(sin54∘)=(IV)sin((cos−1{1√2(cos9π10−sin9π10)}+7π20)−1×π2)(iv)√32(S)If the angles A, B and C of atriangle are in an arithmetic progressionand if a,b and c denote the lengths ofsides opposite to A,B and C respectively,then the value of the expression12(acsin2C+casin2A)is
Which of the following options is only correct combination?
(II), (Q), (iii)
cos(sin−1(sin5π6))=cos(π6)=√32cos(cos−1(−sin7π6))=cos(π3)=12cos(tan−1(tan15π4))=cos(3π4)=−1√2cos−1{1√2(cos9π10−sin9π10−sin9π10)}=cos−1(cos23π20)=2π−23π20=17π20
f(x)=3x3−13x2+14x−2α,β,γα+β+γ=133αβ+βγ+γα=143αβγ=23∴sin(tan−1α+tan−1β+tan−1γ)sin(tan−1(133−231−143))=sin(tan−1(−1))=−1√2
√3cosec20∘−sec20∘=tan60∘cosec20∘−sec20∘=sin60∘cos20∘−cos60∘sin20∘cos60∘.sin20∘cos20∘=sin40∘12sin20∘cos20∘=4(sin12∘)(sin48∘)(sin54∘)=12(2sin12∘sin48∘)sin54∘=12(2sin12∘sin48∘)sin54∘=12[cos36∘−cos60∘]sin54∘=12[cos36∘−12]sin54∘=12[(√5+14)2−12(√5+14)]=18
(iv)Since A,B,C are in A.P. ∠B=60∘
∴ac(2sinCcosC)+ca(2sinAcosA)=2k(acosC+ccosA)=2kb=2sinB=√3