How do you evaluate ${sec}^{- 1} (sec (\frac{7 π}{10}))$ $sec^-1(sec((7pi)/10))$ ?

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How do you evaluate ${sec}^{- 1} (sec (\frac{7 π}{10}))$ $sec^-1(sec((7pi)/10))$ ?

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Answer 1 · 2017-03-26T08:54:20

$\frac{7 π}{10}$ $(7pi)/10$

Explanation:

${sec}^{- 1} x$ $sec^(-1)x$ is defined for x $\in (R -] - 1, 1 [) \to ([0, π] - {\frac{π}{2}})$ $in(R-]-1,1[)rarr([0,pi]-{pi/2})$ .
${sec}^{- 1} sec x = x$ $sec^(-1)secx= x$ if and only if x $\in ([0, π] - {\frac{π}{2}})$ $in([0,pi]-{pi/2})$ .
in your question, since $\frac{7 π}{10} \in ([0, π] - {\frac{π}{2}})$ $(7pi)/10in([0,pi]-{pi/2})$
so, ${sec}^{- 1} sec (\frac{7 π}{10}) = \frac{7 π}{10}$ $sec^(-1)sec((7pi)/10)=(7pi)/10$

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How do you evaluate ${sec}^{- 1} (sec (\frac{7 π}{10}))$ $sec^-1(sec((7pi)/10))$ ?

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How do you evaluate ${sec}^{- 1} (sec (\frac{7 π}{10}))$ $sec^-1(sec((7pi)/10))$ ?