Answer :

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) = .


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