The correct option is C π3
cos−1[−sin(7π6)]=cos−1[cos(π2+7π6)]=cos−1[cos(10π6)]=cos−1[cos(5π3)]=cos−1[cos(2π−π3)]=cos−1[cosπ3]=π3 (∵cos−1(cosx)=x, x∈[0,π])
Alternate solution
cos−1[−sin(7π6)]
=cos−1[−sin(π+π6)]
=cos−1[sin(π6)]
=cos−1(12)
=π3, ∵cos−1(x)∈[0,π]