Question 103305
Let's work this one backwards. 
Imagine a triangle with a hypotenuse of 1. 
{{{drawing( 300, 300, -.2, 1.2, -.2,1.2,
green(line(  0, 0,.87,.6)),
green(line(.87,.6,.87, 0)),
green(line(  0, 0,.87, 0)),
locate(.9,.3,x-1),
locate(.4,.5,1),
locate(.2,.1,A))}}}
From this diagram, 
{{{sin (A) = (x-1)/1}}}
or in other words
{{{A=arcsin(x-1)}}}
Then 
{{{sec(A) = sec(arcsin(x-1))}}}
{{{sec(A) = 1/cos(A)}}}
{{{drawing( 300, 300, -.2, 1.2, -.2,1.2,
green(line(  0, 0,.87,.6)),
green(line(.87,.6,.87, 0)),
green(line(  0, 0,.87, 0)),
locate(.9,.3,x-1),
locate(.4,.5,1),
locate(.6,-.05,y),
locate(.2,.1,A))}}}
Using the new variable y, you can say,
{{{cos(A) = y/1}}}
{{{sec(A) = 1/y}}}
From Pythagorean theorem, 
{{{(x-1)^2+y^2=1}}}
{{{y^2=1-(x-1)^2}}}
{{{y=sqrt(1-(x-1)^2)}}} To be technically correct, it can be + or - {{{sqrt(1-(x-1)^2)}}}
{{{sec(A)=1/sqrt(1-(x-1)^2)}}}
{{{sec(arcsin(x-1))=(+-1)/(sqrt(1-(x-1)^2))}}}