[tex]\cot^4 A - \cot^2 A = 1\\\\\cot^4 A = 1 + \cot^2 A\\\\\frac{\cos^4 A}{\sin^4 A} = 1 + \frac{\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{\sin^2 A}{\sin^2 A}+\frac{\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{\sin^2 A+\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{1}{\sin^2 A}\\\\\cos^4 A = \frac{\sin^4 A}{\sin^2 A}\\\\\cos^4 A = \sin^2 A\\\\[/tex]
This then means,
[tex]\cos^4 A + \cos^2 A = 1\\\\\sin^2 A + \cos^2 A = 1\\\\[/tex]
which is the pythagorean trig identity. This concludes the proof.
Therefore, if [tex]\cot^4 A - \cot^2 A = 1[/tex], then [tex]\cos^4 A + \cos^2 A = 1[/tex]
