Trigonometry Examples

$\frac{\tan^{3} (x) - 1}{\tan (x) - 1} = \sec^{2} (x) + \tan (x)$

Step 1

Start on the left side.

$\frac{\tan^{3} (x) - 1}{\tan (x) - 1}$

Step 2

Simplify.

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Step 2.1

Simplify the numerator.

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Step 2.1.1

Rewrite $1$ as $1^{3}$ .

$\frac{{\tan (x)}^{3} - 1^{3}}{\tan (x) - 1}$

Step 2.1.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = \tan (x)$ and $b = 1$ .

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

Step 2.1.3

Simplify.

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Step 2.1.3.1

Multiply $\tan (x)$ by $1$ .

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

Step 2.1.3.2

One to any power is one.

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

Step 2.2

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

$\tan^{2} (x) + \tan (x) + 1$

Step 3

Apply Pythagorean identity.

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Step 3.1

Move $1$ .

$\tan^{2} (x) + 1 + \tan (x)$

Step 3.2

Apply pythagorean identity.

$\sec^{2} (x) + \tan (x)$

Step 4

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

$\frac{\tan^{3} (x) - 1}{\tan (x) - 1} = \sec^{2} (x) + \tan (x)$ is an identity