Trigonometry Examples

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

Step 1

Move all terms not containing $y$ to the right side of the equation.

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Step 1.1

Add $1$ to both sides of the equation.

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

Step 1.2

Apply pythagorean identity.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\sec^{2} (y) = x$ .

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

Step 3.2

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

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

Step 3.3

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

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Step 3.3.1

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

$\sec (y) = \sqrt{x}$

Step 3.3.2

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

$\sec (y) = - \sqrt{x}$

Step 3.3.3

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

$\sec (y) = \sqrt{x}, - \sqrt{x}$

Step 3.4

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

$\sec (y) = \sqrt{x}$

$\sec (y) = - \sqrt{x}$

Step 3.5

Solve for $y$ in $\sec (y) = \sqrt{x}$ .

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Step 3.5.1

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

$y = arcsec (\sqrt{x})$

Step 3.6

Solve for $y$ in $\sec (y) = - \sqrt{x}$ .

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Step 3.6.1

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

$y = arcsec (- \sqrt{x})$

Step 3.7

List all of the solutions.

$y = arcsec (\sqrt{x})$

$y = arcsec (- \sqrt{x})$

$y = arcsec (\sqrt{x})$

$y = arcsec (- \sqrt{x})$

Step 4

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = arcsec (\sqrt{x}), arcsec (- \sqrt{x})$

Step 5

Verify if $f^{- 1} (x) = arcsec (\sqrt{x}), arcsec (- \sqrt{x})$ is the inverse of $f (x) = \sec^{2} (x)$ .

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Step 5.1

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = \sec^{2} (x)$ and $f^{- 1} (x) = arcsec (\sqrt{x}), arcsec (- \sqrt{x})$ and compare them.

Step 5.2

Find the range of $f (x) = \sec^{2} (x)$ .

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Step 5.2.1

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[1, \infty)$

Step 5.3

Find the domain of $arcsec (\sqrt{x})$ .

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Step 5.3.1

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

Step 5.3.2

Set the argument in $arcsec (\sqrt{x})$ less than or equal to $- 1$ to find where the expression is defined.

$\sqrt{x} \leq - 1$

Step 5.3.3

Solve for $x$ .

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Step 5.3.3.1

To remove the radical on the left side of the inequality, square both sides of the inequality.

${\sqrt{x}}^{2} \leq {(- 1)}^{2}$

Step 5.3.3.2

Simplify each side of the inequality.

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Step 5.3.3.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

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

Step 5.3.3.2.2

Simplify the left side.

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Step 5.3.3.2.2.1

Simplify ${(x^{\frac{1}{2}})}^{2}$ .

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Step 5.3.3.2.2.1.1

Multiply the exponents in ${(x^{\frac{1}{2}})}^{2}$ .

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Step 5.3.3.2.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 5.3.3.2.2.1.1.2

Cancel the common factor of $2$ .

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Step 5.3.3.2.2.1.1.2.1

Cancel the common factor.

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

Step 5.3.3.2.2.1.1.2.2

Rewrite the expression.

$x^{1} \leq {(- 1)}^{2}$

Step 5.3.3.2.2.1.2

Simplify.

$x \leq {(- 1)}^{2}$

Step 5.3.3.2.3

Simplify the right side.

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Step 5.3.3.2.3.1

Raise $- 1$ to the power of $2$ .

$x \leq 1$

Step 5.3.3.3

Find the domain of $\sqrt{x} + 1$ .

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Step 5.3.3.3.1

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

Step 5.3.3.3.2

The domain is all values of $x$ that make the expression defined.

$[0, \infty)$

Step 5.3.3.4

Use each root to create test intervals.

$x < 0$

$0 < x < 1$

$x > 1$

Step 5.3.3.5

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 5.3.3.5.1

Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Step 5.3.3.5.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 5.3.3.5.1.2

Replace $x$ with $- 2$ in the original inequality.

$\sqrt{- 2} \leq - 1$

Step 5.3.3.5.1.3

The left side is not equal to the right side, which means that the given statement is false.

False

Step 5.3.3.5.2

Test a value on the interval $0 < x < 1$ to see if it makes the inequality true.

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Step 5.3.3.5.2.1

Choose a value on the interval $0 < x < 1$ and see if this value makes the original inequality true.

$x = 0.5$

Step 5.3.3.5.2.2

Replace $x$ with $0.5$ in the original inequality.

$\sqrt{0.5} \leq - 1$

Step 5.3.3.5.2.3

The left side $0.70710678$ is greater than the right side $- 1$ , which means that the given statement is false.

False

Step 5.3.3.5.3

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 5.3.3.5.3.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 5.3.3.5.3.2

Replace $x$ with $4$ in the original inequality.

$\sqrt{4} \leq - 1$

Step 5.3.3.5.3.3

The left side $2$ is greater than the right side $- 1$ , which means that the given statement is false.

False

Step 5.3.3.5.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < 0$ False

$0 < x < 1$ False

$x > 1$ False

$x < 0$ False

$0 < x < 1$ False

$x > 1$ False

Step 5.3.3.6

Since there are no numbers that fall within the interval, this inequality has no solution.

No solution

Step 5.3.4

Set the argument in $arcsec (\sqrt{x})$ greater than or equal to $1$ to find where the expression is defined.

$\sqrt{x} \geq 1$

Step 5.3.5

Solve for $x$ .

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Step 5.3.5.1

To remove the radical on the left side of the inequality, square both sides of the inequality.

${\sqrt{x}}^{2} \geq 1^{2}$

Step 5.3.5.2

Simplify each side of the inequality.

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Step 5.3.5.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

${(x^{\frac{1}{2}})}^{2} \geq 1^{2}$

Step 5.3.5.2.2

Simplify the left side.

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Step 5.3.5.2.2.1

Simplify ${(x^{\frac{1}{2}})}^{2}$ .

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Step 5.3.5.2.2.1.1

Multiply the exponents in ${(x^{\frac{1}{2}})}^{2}$ .

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Step 5.3.5.2.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{\frac{1}{2} \cdot 2} \geq 1^{2}$

Step 5.3.5.2.2.1.1.2

Cancel the common factor of $2$ .

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Step 5.3.5.2.2.1.1.2.1

Cancel the common factor.

$x^{\frac{1}{2} \cdot 2} \geq 1^{2}$

Step 5.3.5.2.2.1.1.2.2

Rewrite the expression.

$x^{1} \geq 1^{2}$

Step 5.3.5.2.2.1.2

Simplify.

$x \geq 1^{2}$

Step 5.3.5.2.3

Simplify the right side.

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Step 5.3.5.2.3.1

One to any power is one.

$x \geq 1$

Step 5.3.5.3

Find the domain of $\sqrt{x} - 1$ .

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Step 5.3.5.3.1

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

Step 5.3.5.3.2

The domain is all values of $x$ that make the expression defined.

$[0, \infty)$

Step 5.3.5.4

The solution consists of all of the true intervals.

$x \geq 1$

Step 5.3.6

The domain is all values of $x$ that make the expression defined.

$[0, \infty)$

Step 5.4

Since the domain of $f^{- 1} (x) = arcsec (\sqrt{x}), arcsec (- \sqrt{x})$ is not equal to the range of $f (x) = \sec^{2} (x)$ , then $f^{- 1} (x) = arcsec (\sqrt{x}), arcsec (- \sqrt{x})$ is not an inverse of $f (x) = \sec^{2} (x)$ .

There is no inverse

Step 6