Question 703932
<pre>

{{{(2tan(A))/(1+tan^2(A))}}} = sin(2A)

Work with the left side. Use the identity {{{tan(theta)=sin(theta)/cos(theta)}}}
on the top and use the identity {{{1+tan^2(theta)=sec^2(theta)}}}
on the bottom:

{{{(  2sin(A)/cos(A)  )/(sec^2(A))}}} = 

Use the identity {{{sec(theta)=1/cos(theta)}}} on the bottom:

{{{(  2sin(A)/cos(A)  )/(

(1/cos(A))^2

)}}}

{{{(  2sin(A)/cos(A)  )/(

1/cos^2(A)

)}}}

Invert and multiply:

{{{(  2sin(A)/cos(A)  )*(

cos^2(A)/1

)}}}

Cancel the cosine in the bottom into one of the cosine
factors in the top

{{{(  2sin(A)/cross(cos(A))  )*(

cos^cross(2)(A)/1

)}}}

2sin(A)cos(A)

We use the identity {{{sin(2theta)=2sin(theta)cos(theta)}}}

sin(2A)

Edwin</pre>