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diff --git a/js/src/tests/non262/Math/15.8.2.18.js b/js/src/tests/non262/Math/15.8.2.18.js
new file mode 100644
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+++ b/js/src/tests/non262/Math/15.8.2.18.js
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+/* -*- tab-width: 2; indent-tabs-mode: nil; js-indent-level: 2 -*- */
+/* This Source Code Form is subject to the terms of the Mozilla Public
+ * License, v. 2.0. If a copy of the MPL was not distributed with this
+ * file, You can obtain one at http://mozilla.org/MPL/2.0/. */
+
+
+/**
+   File Name:          15.8.2.18.js
+   ECMA Section:       15.8.2.18 tan( x )
+   Description:        return an approximation to the tan of the
+   argument.  argument is expressed in radians
+   special cases:
+   - if x is NaN           result is NaN
+   - if x is 0             result is 0
+   - if x is -0            result is -0
+   - if x is Infinity or -Infinity result is NaN
+   Author:             christine@netscape.com
+   Date:               7 july 1997
+*/
+
+var SECTION = "15.8.2.18";
+var TITLE   = "Math.tan(x)";
+var EXCLUDE = "true";
+
+writeHeaderToLog( SECTION + " "+ TITLE);
+
+new TestCase( "Math.tan.length",
+	      1,
+	      Math.tan.length );
+
+new TestCase( "Math.tan()",
+	      Number.NaN,
+	      Math.tan() );
+
+new TestCase( "Math.tan(void 0)",
+	      Number.NaN,
+	      Math.tan(void 0));
+
+new TestCase( "Math.tan(null)",
+	      0,
+	      Math.tan(null) );
+
+new TestCase( "Math.tan(false)",
+	      0,
+	      Math.tan(false) );
+
+new TestCase( "Math.tan(NaN)",
+	      Number.NaN,
+	      Math.tan(Number.NaN) );
+
+new TestCase( "Math.tan(0)",
+	      0,
+	      Math.tan(0));
+
+new TestCase( "Math.tan(-0)",
+	      -0,
+	      Math.tan(-0));
+
+new TestCase( "Math.tan(Infinity)",
+	      Number.NaN,
+	      Math.tan(Number.POSITIVE_INFINITY));
+
+new TestCase( "Math.tan(-Infinity)",
+	      Number.NaN,
+	      Math.tan(Number.NEGATIVE_INFINITY));
+
+new TestCase( "Math.tan(Math.PI/4)",
+	      1,
+	      Math.tan(Math.PI/4));
+
+new TestCase( "Math.tan(3*Math.PI/4)",
+	      -1,
+	      Math.tan(3*Math.PI/4));
+
+new TestCase( "Math.tan(Math.PI)",
+	      -0,
+	      Math.tan(Math.PI));
+
+new TestCase( "Math.tan(5*Math.PI/4)",
+	      1,
+	      Math.tan(5*Math.PI/4));
+
+new TestCase( "Math.tan(7*Math.PI/4)",
+	      -1,
+	      Math.tan(7*Math.PI/4));
+
+new TestCase( "Infinity/Math.tan(-0)",
+	      -Infinity,
+	      Infinity/Math.tan(-0) );
+
+/*
+  Arctan (x) ~ PI/2 - 1/x   for large x.  For x = 1.6x10^16, 1/x is about the last binary digit of double precision PI/2.
+  That is to say, perturbing PI/2 by this much is about the smallest rounding error possible.
+
+  This suggests that the answer Christine is getting and a real Infinity are "adjacent" results from the tangent function.  I
+  suspect that tan (PI/2 + one ulp) is a negative result about the same size as tan (PI/2) and that this pair are the closest
+  results to infinity that the algorithm can deliver.
+
+  In any case, my call is that the answer we're seeing is "right".  I suggest the test pass on any result this size or larger.
+  = C =
+*/
+
+new TestCase( "Math.tan(3*Math.PI/2) >= 5443000000000000",
+	      true,
+	      Math.tan(3*Math.PI/2) >= 5443000000000000 );
+
+new TestCase( "Math.tan(Math.PI/2) >= 5443000000000000",
+	      true,
+	      Math.tan(Math.PI/2) >= 5443000000000000 );
+
+test();