用编译器测试并不难。为此，我编写了这个模块...

-module my_mod.
-export [t1/1, t2/1].

-record(test, {a = 10}).

t1(Test) when is_record(Test,test) -> somethings.

t2(#test{} = _Test) -> somethings.

然后，我运行erlc -E my_mod.erl，这是得到的扩展代码：

-file("my_mod.erl", 1).

-module(my_mod).

-export([t1/1,t2/1]).

-record(test,{a = 10}).

t1({test, _} = Test) when true ->
    somethings.

t2({test, _} = _Test) ->
    somethings.

所以，基本上......是一样的。使用is_record(Test, test)会增加一个无用的保护（true），但这不会对 * 速度 * 产生影响。
此外，如果使用erlc -S my_mod.erl生成程序集清单，则会得到：

{module, my_mod}.  %% version = 0

{exports, [{module_info,0},{module_info,1},{t1,1},{t2,1}]}.

{attributes, []}.

{labels, 9}.

{function, t1, 1, 2}.
  {label,1}.
    {line,[{location,"my_mod.erl",6}]}.
    {func_info,{atom,my_mod},{atom,t1},1}.
  {label,2}.
    {test,is_tagged_tuple,{f,1},[{x,0},2,{atom,test}]}.
    {move,{atom,somethings},{x,0}}.
    return.

{function, t2, 1, 4}.
  {label,3}.
    {line,[{location,"my_mod.erl",8}]}.
    {func_info,{atom,my_mod},{atom,t2},1}.
  {label,4}.
    {test,is_tagged_tuple,{f,3},[{x,0},2,{atom,test}]}.
    {move,{atom,somethings},{x,0}}.
    return.

{function, module_info, 0, 6}.
  {label,5}.
    {line,[]}.
    {func_info,{atom,my_mod},{atom,module_info},0}.
  {label,6}.
    {move,{atom,my_mod},{x,0}}.
    {line,[]}.
    {call_ext_only,1,{extfunc,erlang,get_module_info,1}}.

{function, module_info, 1, 8}.
  {label,7}.
    {line,[]}.
    {func_info,{atom,my_mod},{atom,module_info},1}.
  {label,8}.
    {move,{x,0},{x,1}}.
    {move,{atom,my_mod},{x,0}}.
    {line,[]}.
    {call_ext_only,2,{extfunc,erlang,get_module_info,2}}.

如您所见，这两个函数实际上是相同的：

{function, …, 1, …}.
  {label,…}.
    {line,[{location,"my_mod.erl",…}]}.
    {func_info,{atom,my_mod},{atom,…},1}.
  {label,…}.
    {test,is_tagged_tuple,{f,…},[{x,0},2,{atom,test}]}.
    {move,{atom,somethings},{x,0}}.
    return.

也许是第二个，在运行时没有记录的概念，只有元组。

21> rd(test, {a = 10}).               
test
22> erlang:is_record({test, 1}, test).
true