Question 1062923
<pre><b><font size=4>
Given {{{tan(theta) + sec(theta)}}}{{{""=""}}}{{{x}}} show that {{{sin(theta)}}}{{{""=""}}}{{{(x^2-1)/(x^2+1)}}} 

We begin by proving the converse.  That is first we prove:

Given {{{sin(theta)}}}{{{""=""}}}{{{(x^2-1)/(x^2+1)}}} show that {{{tan(theta) + sec(theta)}}}{{{""=""}}}{{{x}}}

[IOW, we start with what we have to prove, prove what was given,
and then we reverse the proof.]

Draw a right triangle containing angle theta with opposite
side x²-1 and hypotenuse x²+1, 

{{{drawing(175,100,-1,6,-1,3,locate(4.2,1.6,x^2-1),locate(1,2,x^2+1),locate(1,.5,theta),
triangle(0,0,4,0,4,2)  )}}}

and we find what the adjacent side 
must equal using the Pythagorean theorem:

{{{c^2=a^2+b^2}}}
{{{(x^2+1)^2=a^2+(x^2-1)^2}}}
{{{(x^2+1)^2-(x^2-1)^2=a^2}}}
factor the left side as the difference of squares:
{{{( (x^2+1)^""-(x^2-1) )( (x^2+1)^""+(x^2-1) )=a^2}}}
{{{( x^2+1-x^2+1 )( x^2+1+x^2-1 )=a^2}}}
{{{(2^"")(2x^2 )=a^2}}}
{{{4x^2 =a^2}}}
{{{2x=a}}}

So we complete the right triangle with adjacent side 2x:

{{{drawing(175,100,-1,6,-1,3,
locate(2,0,2x),locate(4.2,1.6,x^2-1),locate(1,2,x^2+1),locate(1,.5,theta),
triangle(0,0,4,0,4,2)  )}}}

From that triangle we can prove what was given: 

{{{tan(theta) + sec(theta)}}}{{{""=""}}}{{{x}}}

{{{(x^2-1)/(2x)+(x^2+1)/(2x)}}}{{{""=""}}}
{{{(x^2-1+x^2+1)/(2x)}}}{{{""=""}}}
{{{(2x^2)/(2x)}}}{{{""=""}}}
{{{x}}}

So now that we have proved the converse of the problem,
we can reverse the proof:

-----------------------------------

Given {{{tan(theta) + sec(theta)}}}{{{""=""}}}{{{x}}},
reversing the above steps:

{{{x}}}{{{""=""}}}
{{{(2x^2)/(2x)}}}{{{""=""}}}
{{{(x^2-1+x^2+1)/(2x)}}}{{{""=""}}}
{{{(x^2-1)/(2x)+(x^2+1)/(2x)}}}{{{""=""}}}

From that and {{{tan(theta) + sec(theta)}}}{{{""=""}}}{{{x}}}
we can draw the right triangle:

{{{drawing(175,100,-1,6,-1,3,
locate(2,0,2x),locate(4.2,1.6,x^2-1),locate(1,2,x^2+1),locate(1,.5,theta),
triangle(0,0,4,0,4,2)  )}}}

Then from that right triangle, it follows that

{{{sin(theta)}}}{{{""=""}}}{{{(x^2-1)/(x^2+1)}}}

Edwin</pre></b></font>