What is $cos [{sec}^{- 1} (- 5)]$ $cos [Sec ^-1 (-5)]$ ?

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Answer 1 · 2015-10-17T16:20:43

#cos(arcsec(-5)) = -1/5

Knowing that $cos (x) = \frac{1}{sec (x)}$ $cos(x) = 1/sec(x)$ we can say that

$cos (a r c sec (- 5)) = \frac{1}{sec (a r c sec (- 5))}$ $cos(arcsec(-5)) = 1/sec(arcsec(-5))$

Since $sec (a r c sec (x)) = x$ $sec(arcsec(x)) = x$ we have

$cos (a r c sec (- 5)) = \frac{1}{- 5} = - \frac{1}{5}$ $cos(arcsec(-5)) = 1/(-5) = -1/5$

What is cos[sec−1(−5)]cos [Sec ^-1 (-5)]?