Answer :

Let us assume 2tan = θ


We know tan = –1


2tan = 2(–1)


2tan = –2


The question converts to sec–1(–2)


Now,


Let sec–1(–2) = y


sec y = –2


= – sec = 2


= sec


= sec


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


and sec = –2


The principal value of sec–1(2tan) is


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