How do you know how to solve sec 45?

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Answer 1 · 2018-06-21T18:00:50

#sec(45^@)=sqrt2/1=sqrt2#

#sec(45^@)#

#=1/cos(45^@)#

#sin(45^@)=cos(45^@)#

we have an isosceles right triangle
where the ratio of its sides are #1:1:sqrt2#

Cosine = adjacent/hypotenuse

1/cosine=hypotenuse/adjacent
thus:

#sec(45^@)=sqrt2/1=sqrt2#

Answer 2 · 2018-06-22T00:02:08

#sec45=sqrt2#

We know that #secx# is defined as #1/cosx#, so we are essentially being asked to find

#1/cos45#

On the unit circle, the coordinates for #45^@# are #(sqrt2/2,sqrt2/2)#, where the #x#-coordinate is the value for cosine.

This means

#1/cos45=1/(sqrt2/2)#

Which also means that

#sec45=1/(sqrt2/2)#

which simplifies to

#sec45=2/sqrt2#

The convention is to not have an irrational number in the denominator, so we can multiply the top and bottom by #sqrt2#. We get

#sec45=(2sqrt2)/(sqrt2sqrt2)#

#sec45=(cancel2sqrt2)/(cancel2)#

#sec45=sqrt2#

Hope this helps!