hope all are fine

I am using code from following code to use fb like gate on my page.

Code:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<body>
<div id="content">
<div id="Step1">
    <div style="clear:both">
<div>
	<div id="fb-root"></div><script src="//connect.facebook.net/en_US/all.js#appId=209205675779605&amp;xfbml=1"></script><fb:like href="YourPageToLike" send="false" width="208" height="30" show_faces="false" font=""></fb:like>
</div></div></div>
<div id="Step2" style="display: none"></div>
<div id="Step3" style="display: none">
<center><a href="DownloadlinkGoesHere" target="_blank">Click to Download</a></center>
</div>
<div id="fb-root"> </div>

<script type="text/javascript">

window.fbAsyncInit = function() {
    // init the FB JS SDK
    FB.init({
      appId      : '215505698468301', // App ID from the App Dashboard
      status     : true, // check the login status upon init?
      cookie     : true, // set sessions cookies to allow your server to access the session?
      xfbml      : true  // parse XFBML tags on this page?
    });
    FB.Event.subscribe('edge.create', function(href, widget) {
      document.getElementById('Step1').style.display = 'none';
      document.getElementById('Step2').style.display = 'block';
      document.getElementById('Step3').style.display = 'none';
      setTimeout('document.getElementById(\'Step1\').style.display = \'none\';document.getElementById(\'Step2\').style.display = \'none\';document.getElementById(\'Step3\').style.display = \'block\';', 500);
   });
  };
 (function(d, debug){
     var js, id = 'facebook-jssdk', ref = d.getElementsByTagName('script')[0];
     if (d.getElementById(id)) {return;}
     js = d.createElement('script'); js.id = id; js.async = true;
     js.src = "//connect.facebook.net/en_US/all" + (debug ? "/debug" : "") + ".js";
     ref.parentNode.insertBefore(js, ref);
   }(document, /*debug*/ false));
</script>
</div>
</body>
</html>
It works fine, if the user didn't like page, so they can like and get the content. But the problem comes when user already liked the page. So, if he return that page again, like button get grayed/disabled and only way to by pass is unlike and like again

so, what i want: if the page is already liked, the it must show the content instead of shows the like page where like is disabled.

any help would be highly appreciated..

best regards