Trigonometry question please please

Show that cot²θ - cos²θ = cos²θ x cot²θ

Can someone help me with this?
But please don't just give me the answer, I need to know how to do it.
Please explain.
Much appreciated.

Prove: cot²(x) - cos²(x) = cos²(x)cot²(x) \\ x us easier to type than θ

Replace cot(x) = cos(x)/sin(x)

cos²(x)/sin²(x) ‒ cos²(x) = cos²(x)[1/sin²(x) ‒ 1]

= cos²(x)[(1 - sin²(x))/sin²(x)

= cos²(x)[cos²(x)/sin²(x)] = cos²(x)cot²(x)

QED

http://www4a.wolframalpha.com/Calculate/…

cot = cos/sin

cot² - cos² = cos²cot²

cos²/sin² - cos² = cos²cot²

cos²(1/sin² - 1) = cos²cot²

1/sin² - 1 = cot²

1/sin² - 1 = cos²/sin²

Multiply all terms by sin²

1 - sin² = cos²

1 = sin² + cos², which is true