Determine if an iOS device supports TouchID without setting passcode

I'm currently developing an iOS app that enables users to log in to the app using TouchID, but firstly they must set up a password inside the app first. Problem is, to show the setup password option to enable the TouchID login, I need to detect if the iOS device supports TouchID.

Using the LAContext and canEvaluatePolicy (like the answers in here If Device Supports Touch ID), I am able to determine whether the current device supports TouchID if the user has set up passcode on their iOS device. Here is a my code snippet (I'm using Xamarin, so it's in C#):

static bool DeviceSupportsTouchID () 
{
    if (UIDevice.CurrentDevice.CheckSystemVersion(8, 0))
    {
        var context = new LAContext();
        NSError authError;
        bool touchIDSetOnDevice = context.CanEvaluatePolicy(LAPolicy.DeviceOwnerAuthenticationWithBiometrics, out authError);

        return (touchIDSetOnDevice || (LAStatus) Convert.ToInt16(authError.Code) != LAStatus.TouchIDNotAvailable);
    }

    return false;
}

If the user has not set up the device passcode, the authError will just return "PasscodeNotSet" error regardless of whether the device actually supports TouchID or not.

If the user's device supports TouchID, I want to always show the TouchID option in my app regardless of whether the user has set up passcode on their device (I will just warn the user to setup passcode on their device first). Vice versa, if the user's device doesn't support TouchID, I obviously don't want to show the TouchID option in my app.

So my question is, is there a nice way to consistently determine whether an iOS device supports TouchID regardless of whether the user has set up passcode on their device?

The only workaround I can think of is to determine the architecture of the device (which is answered in Determine if iOS device is 32- or 64-bit), as TouchID is only supported on devices with 64-bit architecture. However, I'm looking if there's any nicer way to do this.

Thanks beforehand! :)