Question

Find the principal value of the following:

`sec^-1(2sin  (3pi)/4)`

Answer 1

Let `sec^-1(2sin  (3pi)/4)=y`
Then,
`secy=2sin  (3pi)/4`
We know that the range of the principal value branch is `[0,pi]-{pi/2}.`
Thus,

`secy=2sin  (3pi)/4=2xx1/sqrt2=sqrt2=sec(pi/4)`

`=>y=pi/4in[0,pi]`

Hence, the principal value of `sec^-1(2sin  (3pi)/4)    is    pi/4`