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The principal value of the given equation \mathrm{y=sec^{-1}\sqrt{2}} is

Option: 1

\frac{\pi}{4}


Option: 2

\frac{\pi}{3}


Option: 3

\frac{\pi}{2}


Option: 4

\frac{\pi}{6}


Answers (1)

best_answer

option (c) \frac{\pi}{2}

Given that,

\mathrm{y=sec^{-1}\sqrt{2}}

\mathrm{ sec\ y=\sqrt{2}}

We know that the range of the principal value branch of \mathrm{sec^{-1}x} is \mathrm{ \left ( 0,\pi \right )-\left \{ \frac{\pi}{2} \right \}.}

Thus,

\mathrm{sec\left ( \frac{\pi}{4} \right )={\sqrt{2} }

Therefore, the principal value of   \mathrm{sec^{-1}\left (\sqrt{2} \right )=\frac{\pi}{4}. }

 

 

 

 

 

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mansi

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