Trigonometry Examples

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

Step 1

Start on the left side.

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

Step 2

Simplify.

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Step 2.1

Simplify the denominator.

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Step 2.1.1

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

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

Step 2.1.2

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 $b = \sin (x)$ .

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

Step 2.2

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

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

Step 3

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

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