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 .

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