Find the principal value of sec^{-1}(2 sin( \dfrac{3\pi}{4}))

But we know that \(sin (\dfrac{3\pi}{4}) =  \dfrac{1}{\sqrt{2}}\)

Therefore \( 2sin (\dfrac{3\pi}{4})\)= 2 × \( \dfrac{1}{\sqrt{2}}\)

 \(2sin (\dfrac{3\pi}{4}) =\sqrt{2}\)

Therefore by substituting above values in \( sec^{-1} (2 sin (\dfrac{3\pi}{4}))\), we get

  \( sec^{-1}( \sqrt{2})\)

Let \( sec^{-1}( \sqrt{2})\)= y

Sec y = \( ( \sqrt{2})\)

\(sec (\dfrac{\pi}{4}) =  \sqrt{2}\)

Therefore range of principal value of sec^-1 is [0, π] – {\( \dfrac{\pi}{2}\)}

and  \(sec (\dfrac{\pi}{4}) = \sqrt{2}\)

Thus the principal value of \( sec^{-1} (2 sin (\dfrac{3\pi}{4}))\) is \( \dfrac{\pi}{4}\)