Trigonometry Examples

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

Step 1

Start on the left side.

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

Step 2

Apply Pythagorean identity in reverse.

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

Step 3

Simplify.

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Simplify the numerator.

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

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

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 $b = \cos (x)$ .

$\frac{(1 + \cos (x)) (1 - \cos (x))}{1 - \cos (x)}$

Cancel the common factor of $1 - \cos (x)$ .

$1 + \cos (x)$

Step 4

Rewrite $1 + \cos (x)$ as $\frac{1 + \sec (x)}{\sec (x)}$ .

$\frac{1 + \sec (x)}{\sec (x)}$

Step 5

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

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