<s>[INST] How do I prove this trigonmetric identity? I need to prove that the following identity is true:
$$
\frac{\cos^2x-\sin^2x}{1-\tan^2x}=\cos^2x
$$
This isn't homework; just a practice exercise. But I keep getting stuck! Thanks much.
 [/INST] Let's start with our original expression:
$$\dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}$$
We need to prove that:
$$\dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}=\cos^2 x$$
Step 1: Recall that $\tan x = \dfrac{\sin x}{\cos x}$. This means that $\tan^2 x=\dfrac{\sin^2 x}{\cos^2 x}$.
$$\dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}=\dfrac{\cos^2 x - \sin^2 x}{\left(1-\dfrac{\sin^2 x}{\cos^2 x}\right)}$$
Step 2: Simplify the denominator by letting $1=\dfrac{\cos^2 x}{\cos^2 x}$.
$$\dfrac{\cos^2 x - \sin^2 x}{\left(1-\dfrac{\sin^2 x}{\cos^2 x}\right)}=\dfrac{\cos^2 x - \sin^2 x}{\left(\dfrac{\cos^2 x}{\cos^2 x}-\dfrac{\sin^2 x}{\cos^2 x}\right)}$$
$$\dfrac{\cos^2 x - \sin^2 x}{\left(\dfrac{\cos^2 x}{\cos^2 x}-\dfrac{\sin^2 x}{\cos^2 x}\right)}=\dfrac{\cos^2 x - \sin^2 x}{\left(\dfrac{\cos^2 x-\sin^2 x}{\cos^2 x}\right)}$$
Step 3: Recall that $\dfrac{a}{c}=a\left(\dfrac{1}{c}\right)$.
$$\dfrac{\cos^2 x - \sin^2 x}{\left(\dfrac{\cos^2 x-\sin^2 x}{\cos^2 x}\right)}=\left(\cos^2 x - \sin^2 x\right)\left(\dfrac{\cos^2 x}{\cos^2 x-\sin^2 x}\right)$$
$$\left(\cos^2 x - \sin^2 x\right)\left(\dfrac{\cos^2 x}{\cos^2 x-\sin^2 x}\right)=\cos^2 x\left(\dfrac{\cos^2 x-\sin^2 x}{\cos^2 x-\sin^2 x}\right)$$
$$\cos^2 x\left(\dfrac{\cos^2 x-\sin^2 x}{\cos^2 x-\sin^2 x}\right)=\cos^2 x$$
$$\displaystyle \boxed{\therefore \dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}=\cos^2 x}$$
</s>