Verify :$\cos^2x=\cot^2x-\frac{\cos^2x}{\tan^2x}$

Question

$$\cos^2x=\cot^2x-\frac{\cos^2x}{\tan^2x}$$

How can I solve it?

Answer 1

$$\cos^2x=\frac{\sin^2x}{\tan^2x}\\ \sin^2x=1-\cos^2x\\ \frac{\sin^2x}{\tan^2x}=\frac{1-\cos^2x}{\tan^2x}\\ \frac{1-\cos^2x}{\tan^2x}=\frac{1}{\tan^2x}-\frac{\cos^2x}{\tan^2x}\\ \cos^2x=\cot^2x-\frac{\cos^2x}{\tan^2x}$$

Answer 2

$$\cot^2x-\frac{\cos^2x}{\tan^2x}=\frac{1}{tan^2x}-\frac{\cos^2x}{\tan^2x}=\frac{sin^2x}{tan^2x}=cos^2x $$