How do you prove #Sec(x) - cos(x) = sin(x) * tan(x)#?

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Answer 1 · 2016-07-10T07:59:11

Knowing that #sec(x) = 1/cos(x)# and #tan(x) = sin(x)/cos(x)#, rewriting the equation yields

#1/cos(x) - cos(x)/1 = sin(x) * (sin(x)/cos(x))#

Rewriting the left fraction by using the property

#a/b-c/d =(ad-bc)/(bd)#

and simplifying the right side of the equation yields

#(1-cos^2(x))/cos(x)=(sin^2(x))/cos(x)#

Note that in this case, we can make use of the identity

#sin^2(x)+cos^2(x)=1#, since #1-cos^2(x)=sin^2(x)#, giving us

#sin^2(x)/cos(x)=sin^2(x)/cos(x)#, which is true, therefore we have proven that

#sec(x)-cos(x) = sin(x) tan(x)#