Precalculus Examples

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

Step 1

Start on the right side.

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

Step 2

Factor using the perfect square rule.

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Step 2.1

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

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

Step 2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 \sin (θ) = 2 \cdot 1 \cdot \sin (θ)$

Step 2.3

Rewrite the polynomial.

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

Step 2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 1$ and $b = \sin (θ)$ .

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

Step 3

Apply Pythagorean identity in reverse.

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

Step 4

Simplify.

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Step 4.1

Simplify the denominator.

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Step 4.1.1

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

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

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 $b = \sin (θ)$ .

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

Step 4.2

Cancel the common factor of ${(1 + \sin (θ))}^{2}$ and $1 + \sin (θ)$ .

$\frac{1 + \sin (θ)}{1 - \sin (θ)}$

Step 5

Rewrite $\frac{1 + \sin (θ)}{1 - \sin (θ)}$ as $\frac{\sec (θ) + \tan (θ)}{\sec (θ) - \tan (θ)}$ .

$\frac{\sec (θ) + \tan (θ)}{\sec (θ) - \tan (θ)}$

Step 6

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

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