Answer :

Let,


sin–1 = y


sin y =


–sin y =


–sin


As we know sin(–θ) = –sinθ


–sin = sin


The range of principal value of sin–1 is and sin


Therefore, the principal value of sin–1 is ….(1)


Let us assume 2tan = θ


We know tan


2tan = 2


2tan =


The question converts to sec–1


Now,


Let sec–1 = z


sec z =


= sec


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


and sec


Therefore, the principal value of sec–1(2tan) is …..(2)


Sin–1 – 2sec–1(2tan)


= (from (1) and (2))


=


= –π


Therefore, the value of Sin–1 – 2sec–1(2tan) is –π.


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