Question

Please solve this and only answer with image

Answer 1

$\cot^4 A - \cot^2 A = 1\\\\\cot^4 A = 1 + \cot^2 A\\\\\frac{\cos^4 A}{\sin^4 A} = 1 + \frac{\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{\sin^2 A}{\sin^2 A}+\frac{\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{\sin^2 A+\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{1}{\sin^2 A}\\\\\cos^4 A = \frac{\sin^4 A}{\sin^2 A}\\\\\cos^4 A = \sin^2 A$

This then means,

$\cos^4 A + \cos^2 A = 1\\\\\sin^2 A + \cos^2 A = 1$

which is the pythagorean trig identity. This concludes the proof.

Therefore, if $\cot^4 A - \cot^2 A = 1$, then $\cos^4 A + \cos^2 A = 1$