How do you evaluate ${sec}^{- 1} (sec (\frac{13 π}{10}))$ $sec^-1(sec((13pi)/10))$ ?

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Answer 1 · 2016-10-16T09:34:12

${sec}^{- 1}$ $sec^(-1)$ or $a r c sec$ $arcsec$ is the opposite funtion of $sec$ $sec$ .
They cancel out.

Just like ${\sqrt{x}}^{2} = x and 5 \times x \div 5 = x$ $sqrtx^2=x and 5xx xdiv5=x$

$= \frac{13 π}{10}$ $=(13pi)/10$

How do you evaluate sec−1(sec(13π10))sec^-1(sec((13pi)/10))?