Trigonometry Examples

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

Step 1

Start on the left side.

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

Step 2

Apply Pythagorean identity in reverse.

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

Step 3

Simplify.

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Step 3.1

Simplify each term.

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Step 3.1.1

Simplify the numerator.

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Step 3.1.1.1

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

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

Step 3.1.1.2

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 $b = \cos (x)$ .

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

Step 3.1.2

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

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Step 3.1.2.1

Cancel the common factor.

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

Step 3.1.2.2

Divide $1 + \cos (x)$ by $1$ .

$1 - (1 + \cos (x))$

Step 3.1.3

Apply the distributive property.

$1 - 1 \cdot 1 - \cos (x)$

Step 3.1.4

Multiply $- 1$ by $1$ .

$1 - 1 - \cos (x)$

Step 3.2

Subtract $1$ from $1$ .

$0 - \cos (x)$

Step 3.3

Subtract $\cos (x)$ from $0$ .

$- \cos (x)$

Step 4

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

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