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

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

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