Trigonometry Examples

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

Step 1

Replace the $- \tan^{2} (x)$ with $- (\sec^{2} (x) - 1)$ based on the $\tan^{2} (x) + 1 = \sec^{2} (x)$ identity.

$1 - (\sec^{2} (x) - 1) = \sec^{2} (x)$

Step 2

Simplify each term.

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Step 2.1

Apply the distributive property.

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

Step 2.2

Multiply $- 1$ by $- 1$ .

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

Step 3

Add $1$ and $1$ .

$- \sec^{2} (x) + 2 = \sec^{2} (x)$

Step 4

Move all terms containing $\sec (x)$ to the left side of the equation.

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Step 4.1

Subtract $\sec^{2} (x)$ from both sides of the equation.

$- \sec^{2} (x) + 2 - \sec^{2} (x) = 0$

Step 4.2

Subtract $\sec^{2} (x)$ from $- \sec^{2} (x)$ .

$- 2 \sec^{2} (x) + 2 = 0$

Step 5

Subtract $2$ from both sides of the equation.

$- 2 \sec^{2} (x) = - 2$

Step 6

Divide each term in $- 2 \sec^{2} (x) = - 2$ by $- 2$ and simplify.

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Step 6.1

Divide each term in $- 2 \sec^{2} (x) = - 2$ by $- 2$ .

$\frac{- 2 \sec^{2} (x)}{- 2} = \frac{- 2}{- 2}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $- 2$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{- 2 \sec^{2} (x)}{- 2} = \frac{- 2}{- 2}$

Step 6.2.1.2

Divide $\sec^{2} (x)$ by $1$ .

$\sec^{2} (x) = \frac{- 2}{- 2}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Divide $- 2$ by $- 2$ .

$\sec^{2} (x) = 1$

Step 7

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\sec (x) = \pm \sqrt{1}$

Step 8

Any root of $1$ is $1$ .

$\sec (x) = \pm 1$

Step 9

The complete solution is the result of both the positive and negative portions of the solution.

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Step 9.1

First, use the positive value of the $\pm$ to find the first solution.

$\sec (x) = 1$

Step 9.2

Next, use the negative value of the $\pm$ to find the second solution.

$\sec (x) = - 1$

Step 9.3

The complete solution is the result of both the positive and negative portions of the solution.

$\sec (x) = 1, - 1$

Step 10

Set up each of the solutions to solve for $x$ .

$\sec (x) = 1$

$\sec (x) = - 1$

Step 11

Solve for $x$ in $\sec (x) = 1$ .

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Step 11.1

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

$x = arcsec (1)$

Step 11.2

Simplify the right side.

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Step 11.2.1

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

$x = 0$

Step 11.3

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 11.4

Subtract $0$ from $2 π$ .

$x = 2 π$

Step 11.5

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

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Step 11.5.1

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

$\frac{2 π}{| b |}$

Step 11.5.2

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

$\frac{2 π}{| 1 |}$

Step 11.5.3

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

$\frac{2 π}{1}$

Step 11.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 11.6

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 12

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

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Step 12.1

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

$x = arcsec (- 1)$

Step 12.2

Simplify the right side.

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Step 12.2.1

The exact value of $arcsec (- 1)$ is $π$ .

$x = π$

Step 12.3

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

$x = 2 π - π$

Step 12.4

Subtract $π$ from $2 π$ .

$x = π$

Step 12.5

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

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Step 12.5.1

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

$\frac{2 π}{| b |}$

Step 12.5.2

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

$\frac{2 π}{| 1 |}$

Step 12.5.3

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

$\frac{2 π}{1}$

Step 12.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 12.6

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

$x = π + 2 π n$ , for any integer $n$

Step 13

List all of the solutions.

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

Step 14

Consolidate the answers.

$x = π n$ , for any integer $n$