Verify the Identity (tan(x)^2+1)(cos(x)^2-1)=-tan(x)^2

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Trigonometry

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

Start on the left side.

$(\tan^{2} (x) + 1) (\cos^{2} (x) - 1)$

Apply Pythagorean identity.

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Apply pythagorean identity.

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

Reorder $\cos^{2} (x)$ and $- 1$ .

$\sec^{2} (x) (- 1 + \cos^{2} (x))$

Rewrite $- 1$ as $- 1 (1)$ .

$\sec^{2} (x) (- 1 (1) + \cos^{2} (x))$

Factor $- 1$ out of $\cos^{2} (x)$ .

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

Factor $- 1$ out of $- 1 (1) - 1 (- \cos^{2} (x))$ .

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

Rewrite $- 1 (1 - \cos^{2} (x))$ as $- (1 - \cos^{2} (x))$ .

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

Apply pythagorean identity.

$\sec^{2} (x) (- \sin^{2} (x))$

Convert to sines and cosines.

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Apply the reciprocal identity to $\sec (x)$ .

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

Apply the product rule to $\frac{1}{\cos (x)}$ .

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

Simplify.

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One to any power is one.

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

Combine $\frac{1}{{\cos (x)}^{2}}$ and ${\sin (x)}^{2}$ .

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

Rewrite $- \frac{\sin^{2} (x)}{\cos^{2} (x)}$ as $- \tan^{2} (x)$ .

$- \tan^{2} (x)$

Because the two sides have been shown to be equivalent, the equation is an identity.

$(\tan^{2} (x) + 1) (\cos^{2} (x) - 1) = - \tan^{2} (x)$ is an identity

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