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 .