How do you evaluate $cos [{sec}^{- 1} (- 5)]$ $cos [Sec ^-1 (-5)]$ ?

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Answer 1 · 2016-08-28T02:00:56

$- \frac{1}{5}$ $-1/5$

As cosine is the reciprocal of secant,

$a = {sec}^{- 1} (- 5) = {cos}^{- 1} (- \frac{1}{5})$ $a = sec^(-1)(-5) = cos^(-1)(-1/5)$ ,

the given expression is

$cos a = - \frac{1}{5}$ $cos a = -1/5$ .

Answer 2 · 2016-08-29T02:45:50

$cos ({sec}^{- 1} (- 5)) = - \frac{1}{5}$ $cos(sec^-1(-5))=-1/5$

Let:

$x = cos ({sec}^{- 1} (- 5))$ $x=cos(sec^-1(-5))$

We can then say that:

${cos}^{- 1} (x) = {sec}^{- 1} (- 5)$ $cos^-1(x)=sec^-1(-5)$

Using the same principle to now isolate the $- 5$ $-5$ , we say that:

$sec ({cos}^{- 1} (x)) = - 5$ $sec(cos^-1(x))=-5$

Since $sec (x) = \frac{1}{cos (x)}$ $sec(x)=1/cos(x)$ , rewrite the left-hand side:

$\frac{1}{cos ({cos}^{- 1} (x))} = - 5$ $1/cos(cos^-1(x))=-5$

$cos (x)$ $cos(x)$ and ${cos}^{- 1} (x)$ $cos^-1(x)$ undo one another, being inverse functions:

$\frac{1}{x} = - 5$ $1/x=-5$

Taking the reciprocal of both sides:

$x = - \frac{1}{5}$ $x=-1/5$

Thus:

$cos ({sec}^{- 1} (- 5)) = - \frac{1}{5}$ $cos(sec^-1(-5))=-1/5$

How do you evaluate cos [Sec ^-1 (-5)]cos[sec−1(−5)]cos [Sec ^-1 (-5)]?