How do you know how to solve sec 45?

2 Answers
Jun 21, 2018

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

Explanation:

#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#

Jun 22, 2018

#sec45=sqrt2#

Explanation:

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!