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

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Answer 1 · 2016-08-28T02:00:56

#-1/5#

As cosine is the reciprocal of secant,

# a = sec^(-1)(-5) = cos^(-1)(-1/5)#,

the given expression is

#cos a = -1/5#.

Answer 2 · 2016-08-29T02:45:50

#cos(sec^-1(-5))=-1/5#

Let:

#x=cos(sec^-1(-5))#

We can then say that:

#cos^-1(x)=sec^-1(-5)#

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

#sec(cos^-1(x))=-5#

Since #sec(x)=1/cos(x)#, rewrite the left-hand side:

#1/cos(cos^-1(x))=-5#

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

#1/x=-5#

Taking the reciprocal of both sides:

#x=-1/5#

Thus:

#cos(sec^-1(-5))=-1/5#