# 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.

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$1-\tan^2x=1-\frac{\sin^2x}{\cos^2x}=\frac{\cos^2x-\sin^2x}{\cos^2x}$ – r9m Mar 9 '14 at 0:49
r9m, your answer is the simplest... you should have made it into an answer rather a comment; I would have accepted it! It seems so obvious now. – SoaperGEM Mar 9 '14 at 0:55
it's just an identity :) – r9m Mar 9 '14 at 0:58

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

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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 $\dfrac{a}{c}=a\left(\dfrac{1}{c}\right)$. $$\dfrac{\cos^2 x - \sin^2 x}{\left(\dfrac{\cos^2 x-\sin^2 x}{\cos^2 x}\right)}=\left(\cos^2 x - \sin^2 x\right)\left(\dfrac{\cos^2 x}{\cos^2 x-\sin^2 x}\right)$$ $$\left(\cos^2 x - \sin^2 x\right)\left(\dfrac{\cos^2 x}{\cos^2 x-\sin^2 x}\right)=\cos^2 x\left(\dfrac{\cos^2 x-\sin^2 x}{\cos^2 x-\sin^2 x}\right)$$ $$\cos^2 x\left(\dfrac{\cos^2 x-\sin^2 x}{\cos^2 x-\sin^2 x}\right)=\cos^2 x$$ $$\displaystyle \boxed{\therefore \dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}=\cos^2 x}$$

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Hint: Try factoring $\cos^2 x$ out of the numerator on the left side

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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

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

$$= \ \frac{2 \cos^2 x \ - \ 1 }{2 \ - \ \sec^2 x } \ \cdot \ \frac{\cos^2 x}{\cos^2 x} \ = \ \frac{2 \cos^2 x \ - \ 1 }{ ( 2 - \sec^2 x ) \ \cos^2 x } \ \cdot \ \frac{\cos^2 x}{1}$$

$$= \ \frac{2 \cos^2 x \ - \ 1 }{2 \cos^2 x \ - \ 1 } \ \cdot \ \cos^2 x \ = \ \cos^2 x \ \ .$$

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