Answer :

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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