For the principal values, evaluate the following:
tan–1√3 – sec–1(–2)
The Principal value for tan–1√3
Let tan–1(√3 ) = y
⇒ tan y = √3
The range of principal value of tan–1is {}
And tan = √3
∴ The principal value of tan–1(√3 ) is .
Now,
Principal value for sec–1(–2)
Let sec–1(–2) = z
⇒ sec z = –2
= – sec = 2
= sec
= sec
The range of principal value of sec–1is [0, π]–{}
and sec = –2
Therefore, the principal value of sec–1(–2 ) is .
∴ tan–1√3 –sec–1(–2)
=
=
∴ tan–1√3 – sec–1(–2) = .