RD Sharma - Mathematics (Volume 1)

Book: RD Sharma - Mathematics (Volume 1)

Chapter: 4. Inverse Trigonometric Functions

Subject: Maths - Class 12th

Q. No. 3 of Exercise 4.5

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3

For the principal values, evaluate the following:

Let sec–1(–√2) = y


sec y = –√2


= – sec = √2


= sec


= sec


The range of principal value of sec–1is [0, π]–{}


and sec = –√2.


Let,


cosec–1–√2 = z


cosec z = –√2


–cosec z = √2


–cosec = √2


As we know cosec(–θ) = –cosecθ


–cosec = cosec


The range of principal value of cosec–1 is –{0} and


cosec = –√2


Therefore, the principal value of cosec–1(–√2) is .


cosec–1–√2 = y


cosec y = –√2


–cosec y = √2


–cosec = √2


As we know cosec(–θ) = –cosecθ


–cosec = cosec


The range of principal value of cosec–1 is –{0} and


cosec = –√2


Therefore, the principal value of cosec–1(–√2) is .


From (1) and (2) we get



=


=


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