Q. 2 A5.0( 3 Votes )

# For the principal

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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