Book: RD Sharma - Mathematics (Volume 1)
Chapter: 4. Inverse Trigonometric Functions
Subject: Maths - Class 12th
Q. No. 2 of Exercise 4.4
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2
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) = 
.
