Precalculus Examples

$(\tan (x) - 1) (\sec (x) - 1) = 0$

Step 1

Simplify the left side of the equation.

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Step 1.1

Expand $(\tan (x) - 1) (\sec (x) - 1)$ using the FOIL Method.

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Step 1.1.1

Apply the distributive property.

$\tan (x) (\sec (x) - 1) - 1 (\sec (x) - 1) = 0$

Step 1.1.2

Apply the distributive property.

$\tan (x) \sec (x) + \tan (x) \cdot - 1 - 1 (\sec (x) - 1) = 0$

Step 1.1.3

Apply the distributive property.

$\tan (x) \sec (x) + \tan (x) \cdot - 1 - 1 \sec (x) - 1 \cdot - 1 = 0$

Step 1.2

Simplify each term.

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Step 1.2.1

Move $- 1$ to the left of $\tan (x)$ .

$\tan (x) \sec (x) - 1 \cdot \tan (x) - 1 \sec (x) - 1 \cdot - 1 = 0$

Step 1.2.2

Rewrite $- 1 \tan (x)$ as $- \tan (x)$ .

$\tan (x) \sec (x) - \tan (x) - 1 \sec (x) - 1 \cdot - 1 = 0$

Step 1.2.3

Rewrite $- 1 \sec (x)$ as $- \sec (x)$ .

$\tan (x) \sec (x) - \tan (x) - \sec (x) - 1 \cdot - 1 = 0$

Step 1.2.4

Multiply $- 1$ by $- 1$ .

$\tan (x) \sec (x) - \tan (x) - \sec (x) + 1 = 0$

Step 2

Factor $\tan (x) \sec (x) - \tan (x) - \sec (x) + 1$ .

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Step 2.1

Factor out the greatest common factor from each group.

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Step 2.1.1

Group the first two terms and the last two terms.

$\tan (x) \sec (x) - \tan (x) - \sec (x) + 1 = 0$

Step 2.1.2

Factor out the greatest common factor (GCF) from each group.

$\tan (x) (\sec (x) - 1) - (\sec (x) - 1) = 0$

Step 2.2

Factor the polynomial by factoring out the greatest common factor, $\sec (x) - 1$ .

$(\sec (x) - 1) (\tan (x) - 1) = 0$

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\sec (x) - 1 = 0$

$\tan (x) - 1 = 0$

Step 4

Set $\sec (x) - 1$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $\sec (x) - 1$ equal to $0$ .

$\sec (x) - 1 = 0$

Step 4.2

Solve $\sec (x) - 1 = 0$ for $x$ .

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Step 4.2.1

Add $1$ to both sides of the equation.

$\sec (x) = 1$

Step 4.2.2

Take the inverse secant of both sides of the equation to extract $x$ from inside the secant.

$x = arcsec (1)$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

The exact value of $arcsec (1)$ is $0$ .

$x = 0$

Step 4.2.4

The secant function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$x = 2 π - 0$

Step 4.2.5

Subtract $0$ from $2 π$ .

$x = 2 π$

Step 4.2.6

Find the period of $\sec (x)$ .

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Step 4.2.6.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 4.2.6.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 4.2.6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 4.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.2.7

The period of the $\sec (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = 2 π n, 2 π + 2 π n$ , for any integer $n$

Step 5

Set $\tan (x) - 1$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $\tan (x) - 1$ equal to $0$ .

$\tan (x) - 1 = 0$

Step 5.2

Solve $\tan (x) - 1 = 0$ for $x$ .

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Step 5.2.1

Add $1$ to both sides of the equation.

$\tan (x) = 1$

Step 5.2.2

Take the inverse tangent of both sides of the equation to extract $x$ from inside the tangent.

$x = \arctan (1)$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$x = \frac{π}{4}$

Step 5.2.4

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$x = π + \frac{π}{4}$

Step 5.2.5

Simplify $π + \frac{π}{4}$ .

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Step 5.2.5.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x = π \cdot \frac{4}{4} + \frac{π}{4}$

Step 5.2.5.2

Combine fractions.

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Step 5.2.5.2.1

Combine $π$ and $\frac{4}{4}$ .

$x = \frac{π \cdot 4}{4} + \frac{π}{4}$

Step 5.2.5.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 + π}{4}$

Step 5.2.5.3

Simplify the numerator.

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Step 5.2.5.3.1

Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π + π}{4}$

Step 5.2.5.3.2

Add $4 π$ and $π$ .

$x = \frac{5 π}{4}$

Step 5.2.6

Find the period of $\tan (x)$ .

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Step 5.2.6.1

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 5.2.6.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 5.2.6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 5.2.6.4

Divide $π$ by $1$ .

$π$

Step 5.2.7

The period of the $\tan (x)$ function is $π$ so values will repeat every $π$ radians in both directions.

$x = \frac{π}{4} + π n, \frac{5 π}{4} + π n$ , for any integer $n$

Step 6

The final solution is all the values that make $(\sec (x) - 1) (\tan (x) - 1) = 0$ true.

$x = 2 π n, 2 π + 2 π n, \frac{π}{4} + π n, \frac{5 π}{4} + π n$ , for any integer $n$

Step 7

Consolidate the answers.

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Step 7.1

Consolidate $2 π n$ and $2 π + 2 π n$ to $2 π n$ .

$x = 2 π n, \frac{π}{4} + π n, \frac{5 π}{4} + π n$ , for any integer $n$

Step 7.2

Consolidate $\frac{π}{4} + π n$ and $\frac{5 π}{4} + π n$ to $\frac{π}{4} + π n$ .

$x = 2 π n, \frac{π}{4} + π n$ , for any integer $n$