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Find the principal values of the following: 7. sec inverse of 2 / root 3

Find the principal values of the following:

    7. \sec^{-1}\left (\frac{2}{\sqrt3}\right)

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Let us assume that \sec^{-1}\left (\frac{2}{\sqrt3}\right) = z  then,

we can also write it as; \sec z = \left (\frac{2}{\sqrt3}\right).

Or \sec (\frac{\pi}{6}) = \left (\frac{2}{\sqrt3}\right)  and the principal values lies between  \left [ 0, \pi \right ] - \left \{ \frac{\pi}{2} \right \}.

Hence we get only one principal value of \sec^{-1}\left (\frac{2}{\sqrt3}\right)  i.e., \frac{\pi}{6}.

 

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