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For the principal values, evaluate the following:
First of all we need to find the principal value for cosec–1(–2)
Let,
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 .
∴ Now, the question changes to
Sin–1[cos]
Cos(–θ) = cos(θ)
∴ we can write the above expression as
Sin–1[cos]
Let,
Sin–1 = y
⇒ sin y =
⇒ sin
The range of principal value of sin–1 is and sin
Therefore, the principal value of Sin–1 is
.
Hence, the principal value of the given equation is .