Answer :

Let us assume 2sin = θ

We know sin


2sin = 2


2sin = √2


The question becomes sec–1(√2)


Now,


Let sec–1(√2) = y


sec y = √2


sec = √2


The range of principal value of sec–1is [0, π ]–{}


And sec = √2


The principal value of sec–1(2sin) is .


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