Find the principal value of \( sec^{-1} (-\sqrt{2)}\)

Asked by Sakshi | 2 years ago |  121

1 Answer

Solution :-

Given sec-1 \( (-\sqrt{2})\)

Now let y = sec-1 \( (-\sqrt{2})\)

Sec y = \( -\sqrt{2}\)

We know that  \( sec \dfrac{\pi}{4} =\sqrt{2}\)

Therefore,  \( -sec (\dfrac{\pi}{4}) =- \sqrt{2}\)

= \( sec (π – \dfrac{\pi}{4})\)

= \( sec (\dfrac{3\pi}{4})\)

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

And \( sec (\dfrac{3\pi}{4})=- \sqrt{2}\)

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

Answered by Aaryan | 2 years ago

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