(i) Given as sec^-1 (-√2)

Let y = sec^-1 (-√2)

sec y = -√2

As we know sec π/4 = √2

So, – sec (π/4) = √2

= sec (π – π/4)

= sec (3π/4)

So, the range of principal value of sec^-1 is [0, π] – {π/2}

And sec(3π/4) = – √2

Thus, the principal value of sec^-1 (-√2) is 3π/4

(ii) Given as sec^-1 (2)

Let y = sec^-1 (2)

sec y = 2

= sec π/3

So, the range of principal value of sec^-1 is [0, π] – {π/2} and sec π/3 = 2

Hence, the principal value of sec^-1 (2) is π/3

(iii) Given sec^-1 (2 sin(3π/4))

As we know that sin(3π/4) = 1/√2

So, 2 sin(3π/4) = 2 × 1/√2

2 sin(3π/4) = √2

So, by substituting above values in sec^-1 (2 sin(3π/4)), we get

sec^-1 (√2)

Let sec^-1 (√2) = y

sec y = √2

sec (π/4) = √2

So, range of principal value of sec^-1 is [0, π] – {π/2} and sec (π/4) = √2

Hence, the principal value of sec^-1 (2 sin(3π/4)) is π/4.

(iv) Given as sec^-1 (2 tan(3π/4))

As we know that tan(3π/4) = -1

So, 2 tan(3π/4) = 2 × -1

2 tan (3π/4) = -2

Substitute these values in sec^-1 (2 tan(3π/4)), we get

sec^-1 (-2)

Let y = sec^-1 (-2)

sec y = – 2

– sec(π/3) = 2

= sec(π – π/3)

= sec(2π/3)

So, the range of principal value of sec^-1 is [0, π] – {π/2} and sec (2π/3) = -2

Hence, the principal value of sec^-1 (2 tan(3π/4)) is (2π/3)