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

Question

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

Also, we have not learned the massive list of identities. We were taught to derive everything. I don't know if this necessarily matters, just an FYI.

Answer 1

Hint: The left-hand side is given by $$\frac{\cos^2(x)-\sin^2(x)}{1-\frac{\sin^2(x)}{\cos^2(x)}}=\frac{\cos^2(x)-\sin^2(x)}{\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}}=…$$

Answer 2

$$1-\tan^2{(x)}=1-\frac{\sin^2{(x)}}{\cos^2{(x)}}=\frac1{\cos^2{(x)}}(\cos^2{(x)}-\sin^2{(x)})$$ $$\therefore \frac{\cos^2{(x)}-\sin^2{(x)}}{1-\tan^2{(x)}}=\frac1{\left(\frac1{\cos^2{(x)}}\right)}=\cos^2{(x)}$$

Answer 3

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

This means that $\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}=1-\tan^2(x)$

The left hand side simplfied to $1-\tan^2(x)$, which is the right hand side.