For the principal values, evaluate the following:

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^{–1}is [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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