Find the value of sec−1(−2)−sin−1(21)

Knowing that the range of sec−1x is [0,2π)∪(2π,π] and that of sin−1x is [2π,2π],sec−1(−2)=π−sec−1(2)=32πsin−1(21)=6π∴ sec−1(−2)−sin−1(21) =32π−6π=2π

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Find the value of ^{-1} (-2) - ^{-1} ( {1}{2} )

Question

Find the value of $sec^{- 1} (- 2) - sin^{- 1} (\frac{1}{2})$

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Find the value of

sec^{- 1} (- 2) - sin^{- 1} (\frac{1}{2})

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8 mins ago

Knowing that the range of

se c^{- 1} x

[0, \frac{π}{2}) \cup (\frac{π}{2}, π]

and that of

s i n^{- 1} x

[\frac{π}{2}, \frac{π}{2}]

se c^{- 1} (- 2) = π - se c^{- 1} (2) = \frac{2 π}{3}

s i n^{- 1} (\frac{1}{2}) = \frac{π}{6}

∴

se c^{- 1} (- 2) - s i n^{- 1} (\frac{1}{2})

= \frac{2 π}{3} - \frac{π}{6} = \frac{π}{2}

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