Question 1153180
If arctan(x)=arccos(y), show that 
{{{y =1/sqrt(1+x^2)}}}
<pre>
Let the angle that both sides equal be θ: 

{{{theta=arctan(x/1)=arccos(y/1)}}}

So

{{{matrix(1,3,tan(theta)=x/1,and,cos(theta)=y/1)}}}

and since 

{{{matrix(1,3,tangent=opposite/adjacent, and, cosine=adjacent/hypotenuse)}}},

we can draw two similar right triangles with angle θ, one
for each side of the equation.

{{{drawing(400,150,-1,15,-1,5,

triangle(0,0,4,0,4,3), triangle(7,0,12,0,12,3.75),
locate(2,0,1), locate(4.1,1.5,x), locate(9.5,0,y), locate(9.2,2.4,1),
locate(1,.7,theta), locate(8,.7,theta)  )}}}

We use the Pythagorean theorem to find the missing sides:

{{{drawing(400,150,-1,15,-1,5,

triangle(0,0,4,0,4,3), triangle(7,0,12,0,12,3.75),
locate(2,0,1), locate(4.1,1.5,x), locate(9.5,0,y), locate(9.2,2.4,1),
locate(1,.7,theta), locate(8,.7,theta),
locate(.3,2.5,sqrt(1+x^2)), locate(12,2.3,sqrt(1-y^2))



  )}}}

Because they are similar right triangles,

{{{matrix(1,3,

matrix(1,5,BOTTOM,SIDE,OF,SECOND,TRIANGLE)/matrix(1,5,BOTTOM,SIDE,OF,FIRST,TRIANGLE),
""="",
matrix(1,4,HYPOTENUSE,OF,SECOND,TRIANGLE)/matrix(1,4,HYPOTENUSE,OF,FIRST,TRIANGLE) 
)}}}

{{{matrix(1,3,y^""/1^"",
""="",
1^""/sqrt(1+x^2))}}}

{{{matrix(1,3,y,
""="",
1^""/sqrt(1+x^2))}}}

Edwin</pre>