How to tell the filter views is using
Adding an exposed view filter inserts the following (sample) code into the view output:

 &lt;div class="view view-... view-id-... view-display-id-default view-...-id-1"&gt;
    &lt;div class="view-header"&gt;
        ...
    &lt;/div&gt;

    &lt;div class="view-filters"&gt;
        &lt;form action="..." accept-charset="UTF-8" method="get" id="views-exposed-form-..."&gt;
        &lt;div&gt;
            &lt;div class="views-exposed-form"&gt;
               &lt;div class="views-exposed-widgets clear-block"&gt;
                  &lt;div class="views-exposed-widget views-widget-filter-tid"&gt;
                     &lt;label for="edit-tid"&gt;...&lt;/label&gt;
                     &lt;div class="views-widget"&gt;
                        &lt;div class="form-item" id="edit-tid-wrapper"&gt;
                           &lt;select name="tid" class="form-select" id="edit-tid" &gt;
                              &lt;option value="All"&gt;...&lt;/option&gt;...
                           &lt;/select&gt;
                        &lt;/div&gt;
                     &lt;/div&gt;
                 &lt;/div&gt;
                 &lt;div class="views-exposed-widget views-submit-button"&gt;
                    &lt;input type="submit" id="edit-submit-..." class="form-submit" /&gt;
                 ...


It is not clear, though, why you can't find out the same by simply creating a view, exposing its filter and looking up the source code. Perhaps you could explain your question better?