How do you evaluate ${sec}^{- 1} (sec (\frac{9 π}{10}))$ $sec^-1(sec((9pi)/10))$ ?

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How do you evaluate ${sec}^{- 1} (sec (\frac{9 π}{10}))$ $sec^-1(sec((9pi)/10))$ ?

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Answer 1 · 2018-07-04T15:59:39

${sec}^{- 1} (sec (\frac{9 π}{10})) = \frac{9 π}{10}$ $sec^-1(sec((9pi)/10))=(9pi)/10$

Explanation:

The range of ${sec}^{- 1}$ $sec^-1$ is ${sec}^{- 1} \in [0, \frac{π}{2}) \cup (\frac{π}{2}, π]$ $sec^-1in[0,pi/2)uu(pi/2,pi]$ and since $\frac{9 π}{10} \in [0, \frac{π}{2}) \cup (\frac{π}{2}, π]$ $(9pi)/10in[0,pi/2)uu(pi/2,pi]$ , ${sec}^{- 1} (sec (\frac{9 π}{10})) = \frac{9 π}{10}$ $sec^-1(sec((9pi)/10))=(9pi)/10$ .

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How do you evaluate ${sec}^{- 1} (sec (\frac{9 π}{10}))$ $sec^-1(sec((9pi)/10))$ ?

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How do you evaluate sec−1(sec(9π10))sec^-1(sec((9pi)/10))?

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How do you evaluate ${sec}^{- 1} (sec (\frac{9 π}{10}))$ $sec^-1(sec((9pi)/10))$ ?