 ## 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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##### 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 = = 1
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