Precalculus Examples

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

Step 1

Start on the left side.

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

Step 2

Simplify.

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Step 2.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \sin (x)$ and $b = \cos (x)$ .

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

Step 2.2

Cancel the common factor of ${(\sin (x) + \cos (x))}^{2}$ and $\sin (x) + \cos (x)$ .

$\frac{\sin (x) + \cos (x)}{\sin (x) - \cos (x)}$

Step 3

Reorder terms.

$\frac{\cos (x) + \sin (x)}{\sin (x) - \cos (x)}$

Step 4

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

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

Step 5

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

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