Python requests same header like chrome but different response (show Cloudflare) - python-3.x

Question

I'm trying to request (GET Method) with requests Library But I don't know why I get different response from this URL (udemy.com). can be problem from Certificate or cipher or Protocol?
headers = {
'Connection': 'keep-alive',
'Upgrade-Insecure-Requests': '1',
'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/98.0.4758.82 Safari/537.36',
'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.9',
'Accept-Language': 'en-US,en;q=0.9',
'Accept-Encoding': 'gzip, deflate',
}
req1 = requests.get('https://www.udemy.com/join/signup-popup/?displayType=ajax&display_type=popup&returnUrlAfterLogin=https&showSkipButton=1',headers=headers)
print(req1.text)
<Response [403]>

Answer 1

Cloudflare could detect pretty much every script, but if you use the browser itself with selenium there should be no problem.
from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.firefox.options import Options
try:
options = Options()
options.headless = True # for background run
browser = webdriver.Firefox(options=options) # required for firefox: https://github.com/mozilla/geckodriver/releases
browser.get('https://www.udemy.com/join/signup-popup/?displayType=ajax&display_type=popup&returnUrlAfterLogin=https&showSkipButton=1')
print(browser.find_element(By.ID, "auth-to-udemy-title").text) # Prints "Sign Up and Start Learning!"
finally:
if browser:
browser.close() # avoid memory leakage
The Selenium package and for Firefox the geckodriver are required.

Answer 2

The site likely does not want scripts to access it, and is probably using some sort of bot detection to block them. Trying to work around this would be unethical, and possibly illegal. The ethical way to proceed is to ask the site owner for permission to use your script, and have them give you some sort of bypass token for that purpose.