Precalculus Examples

$\frac{\sin^{2} (θ)}{\cos^{2} (θ)} + 1 = \sec^{2} (θ)$

Step 1

Start on the left side.

$\frac{\sin^{2} (θ)}{\cos^{2} (θ)} + 1$

Step 2

Apply Pythagorean identity in reverse.

$\frac{1 - \cos^{2} (θ)}{\cos^{2} (θ)} + 1$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Rewrite $1$ as $1^{2}$ .

$\frac{1^{2} - {\cos (θ)}^{2}}{{\cos (θ)}^{2}} + 1$

Step 3.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = \cos (θ)$ .

$\frac{(1 + \cos (θ)) (1 - \cos (θ))}{{\cos (θ)}^{2}} + 1$

Step 3.2

Write $1$ as a fraction with a common denominator.

$\frac{(1 + \cos (θ)) (1 - \cos (θ))}{{\cos (θ)}^{2}} + \frac{{\cos (θ)}^{2}}{{\cos (θ)}^{2}}$

Step 3.3

Combine the numerators over the common denominator.

$\frac{(1 + \cos (θ)) (1 - \cos (θ)) + {\cos (θ)}^{2}}{{\cos (θ)}^{2}}$

Step 3.4

Simplify the numerator.

$\frac{1}{\cos^{2} (θ)}$

Step 4

Rewrite $\frac{1}{\cos^{2} (θ)}$ as $\sec^{2} (θ)$ .

$\sec^{2} (θ)$

Step 5

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

$\frac{\sin^{2} (θ)}{\cos^{2} (θ)} + 1 = \sec^{2} (θ)$ is an identity