Question 212096
Note: In order to use the Algebra.com's software more effectively, I am going to use "x" instead of "θ".

{{{(sin(90-x)/sec(90- x))*(tan(90- x)/cos(x)) = cos(x)}}}<br>
Often the easiest way to prove identities is to use basic identities to convert all other trig functions into sin and/or cos. So we will start by replacing sec and tan using {{{sec(x) = 1/cos(x)}}} and {{{tan(x) = sin(x)/cos(x)}}}:<br>
{{{(sin(90-x)/(1/cos(90- x)))*(((sin(90- x))/(cos(90- x)))/cos(x)) = cos(x)}}}
To simplify this compound fraction, all we need to do is multiply the top and bottom of the "big" fraction by cos(90- x):
{{{((cos(90- x))/(cos(90- x)))*(sin(90-x)/(1/cos(90- x)))*(((sin(90- x))/(cos(90- x)))/cos(x)) = cos(x)}}}
Now we can cancel the cos(90-x)'s:
{{{((red(cross(cos(90- x))))/(green(cross(cos(90- x)))))*(sin(90-x)/(1/green(cross(cos(90- x)))))*(((sin(90- x))/(red(cross(cos(90- x)))))/cos(x)) = cos(x)}}}
leaving
{{{(sin(90-x)*sin(90-x))/cos(x) = cos(x)}}}
Since sin(90-x) = cos(x) we can substitute:
{{{(cos(x)*cos(x))/cos(x) = cos(x)}}}
We can cancel a cos(x):
{{{(cross(cos(x))*cos(x))/cross(cos(x)) = cos(x)}}}
leaving
{{{cos(x) = cos(x)}}}<br>
Note: This is just one way to prove this identity. There are many others.