How to divide trigonometric ratios using identities? $$\frac{1-\tan^2x}{1+\tan^2x}$$
We know:
$$\frac{1-\frac{\sin^2x}{\cos^2x}}{1+\frac{\sin^2x}{\cos^2x}}$$
Now what? Flip denominator and times numerator?
Which equals ???
Please help - Thanks

$$
\tan^2 x + 1 = \sec^2 x
$$
Thus
$$
\frac{1-\tan^2 x}{\sec^2 x} = \cos^2x - \sin^2x 
$$
But going from your point of view
$$
\frac{1-\frac{\sin^2x}{\cos^2x}}{1+\frac{\sin^2x}{\cos^2x}}= \frac{\frac{1}{\cos^2x}}{\frac{1}{\cos^2x}}\frac{\cos^2-\sin^2x}{\cos^2x+\sin^2x}=\\
\frac{\cos^2-\sin^2x}{\cos^2x+\sin^2x}
$$
then use the fact
$$
\cos^2x+\sin^2x=1
$$
Dividing through by $\cos^2x$ and knowing that $\frac{1}{\cos x} =\sec x$ will lead to the first identity.
