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.
 A: $\cos^2(x)(1-\tan^2(x))
=\cos^2(x)- \cos^2(x)\frac{\sin^2(x)}{\cos^2(x)}
=\cos^2(x)-\sin^2(x)
$.
A: 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 $\dfrac{a}{c}=a\left(\dfrac{1}{c}\right)$.
$$\dfrac{\cos^2 x - \sin^2 x}{\left(\dfrac{\cos^2 x-\sin^2 x}{\cos^2 x}\right)}=\left(\cos^2 x - \sin^2 x\right)\left(\dfrac{\cos^2 x}{\cos^2 x-\sin^2 x}\right)$$
$$\left(\cos^2 x - \sin^2 x\right)\left(\dfrac{\cos^2 x}{\cos^2 x-\sin^2 x}\right)=\cos^2 x\left(\dfrac{\cos^2 x-\sin^2 x}{\cos^2 x-\sin^2 x}\right)$$
$$\cos^2 x\left(\dfrac{\cos^2 x-\sin^2 x}{\cos^2 x-\sin^2 x}\right)=\cos^2 x$$
$$\displaystyle \boxed{\therefore \dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}=\cos^2 x}$$
A: Hint: Try factoring $\cos^2 x$ out of the numerator on the left side
A: This is sort of the "reverse" of  MPW's suggestion, using the difference of two squares:
$$ \frac{\cos^2 x \ - \ \sin^2 x}{1-\tan^2x} \ = \ \frac{(\cos x - \sin x) \ (\cos x + \sin x) }{(1 - \tan x) \ (1 + \tan x)}  $$
$$ = \ \frac{(\cos x - \sin x) \ (\cos x + \sin x) }{(1 - \tan x) \ (1 + \tan x)} \ \cdot \ \frac{\cos^2 x}{\cos^2 x} $$
$$ = \ \frac{(\cos x - \sin x) \ (\cos x + \sin x) }{\cos x \cdot (1 - \tan x) \cdot \cos x \cdot (1 + \tan x)} \ \cdot \ \frac{\cos^2 x}{1}  $$  
$$= \ \frac{(\cos x - \sin x) \ (\cos x + \sin x) }{(\cos x - \sin x) \ (\cos x + \sin x)} \ \cdot \ \cos^2 x \ = \ \cos^2x \ \ . $$
We could also use two versions of the Pythagorean Identity to write
$$ \frac{\cos^2 x \ - \ \sin^2 x}{1 \ - \ \tan^2x} \ = \ \frac{\cos^2 x \ - \ [  1 - \cos^2 x]}{1 - \ [\sec^2 x - 1]} \ = \ \frac{2 \cos^2 x \ - \   1 }{2  \ - \ \sec^2 x }  $$
$$ = \ \frac{2 \cos^2 x \ - \   1 }{2  \ - \ \sec^2 x  } \ \cdot \ \frac{\cos^2 x}{\cos^2 x} \  = \ \frac{2 \cos^2 x \ - \   1 }{ ( 2   -  \sec^2 x ) \ \cos^2 x  } \ \cdot \ \frac{\cos^2 x}{1}  $$
$$ = \ \frac{2 \cos^2 x \ - \   1 }{2 \cos^2 x \ - \   1 } \ \cdot \ \cos^2 x \ = \  \cos^2 x \ \ .  $$
