Find the principal value of each of the following:
i) `sin^(-1)(1/2)`
ii) `sin^(-1)(-sqrt(3)/2)`
iii) `tan^(-1)(-1/sqrt(3))`
iv) `tan^(-1)(-1)`
v) `"cosec"^(-1)(-2)`
vi) `tan^(-1)(-sqrt(3))`

Explanation

i) We know that the principal-value branch of `sin^(-1)` is `[-pi/2, pi/2]`.
Let `sin^(-1)(-1/2)=theta`. Then,
`sintheta=-1/2=sin(-pi/6)`, where `-pi/6 in [-pi/2, pi/2]`.
Hence, the principal value of `sin^(-1)(-1/2)` is `-pi/6`.
ii) We know that the principal-value branch of `sin^(-1)` is `[-pi/2,pi/2]`.
Let `sin^(-1)(-sqrt(3)/2)=theta`. Then,
`sintheta=-sqrt(3)/2=sin(-pi/3) in [-pi/2,pi/2]`.
Hence, the principal value of `sin^(-1)(-sqrt(3)/2)` is `-pi/3`
iii) We know that the principal-value branch of `tan^(-1)` is `(-pi/2, pi/2)`.
Let `tan^(-1)(-1/sqrt(3))=theta`. Then, `tantheta=-1/sqrt(3)=tan(-pi/6)`, where `-pi/6 in (-pi/2, pi/2)`.
Hence, the principal value of `tan^(-1)(-1/sqrt(3))` is `-pi/6`.
iv) We know that the principal-value branch of `tan^(-1)` is `(-pi/2, pi/2)`.
Let `tan^(-1)(-1)=theta`. Then,
`tantheta(-1)=theta`. Then,
`tantheta=-1=tan(-pi/4)`, where `-pi/4 in (-pi/2, pi/2)`.
Hence, the principal value of `tan^(-1)(-1)` is `-pi/4`.
v) We know that the princpal-value branch of `"cosec"^(-1)` is `[-pi/2, pi/2]-{0}`.
Let `"cosec^(-1)(-2)=theta`. Then,
`"cosec"theta=-2="cosec"(-pi/6)`, where `-pi/6 in [-pi/2, pi/2]-{0}`.
Hence, the principal value of `"cosec"^(-1)(-2)` is `-pi/6`.
vi) We know that the principal-value branch of `tan^(-1)` is `(-pi/2, pi/2)`.
Let `tan^(-1)(-sqrt(3))=theta`. Then,
`tantheta=-sqrt(3)=tan(-pi/3)`, where `-pi/3 in (-pi/2, pi/2)`.
Hence, the principal value of `tan^(-1)(-sqrt(3))` is `-pi/3`.