Precalculus Examples

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

Step 1

Start on the left side.

$1 + \sec^{2} (θ) \cdot \sin^{2} (θ)$

Step 2

Simplify each term.

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Step 2.1

Rewrite $\sec (θ)$ in terms of sines and cosines.

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

Step 2.2

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

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

Step 2.3

One to any power is one.

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

Step 2.4

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

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

Step 3

Apply Pythagorean identity in reverse.

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

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

Step 4.1.2

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 $b = \cos (θ)$ .

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

Step 4.2

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

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

Step 4.3

Combine the numerators over the common denominator.

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

Step 4.4

Simplify the numerator.

$\frac{1}{\cos^{2} (θ)}$

Step 5

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

$\sec^{2} (θ)$

Step 6

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

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