Q. 2 B

# For the principal

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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