Trigonometry Examples

$\tan (\arccos (6 x))$

Step 1

Interchange the variables.

$x = \tan (\arccos (6 y))$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\tan (\arccos (6 y)) = x$ .

$\tan (\arccos (6 y)) = x$

Step 2.2

Take the inverse tangent of both sides of the equation to extract $\arccos (6 y)$ from inside the tangent.

$\arccos (6 y) = \arctan (x)$

Step 2.3

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

$6 y = \cos (\arctan (x))$

Step 2.4

Simplify the right side.

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Step 2.4.1

Simplify $\cos (\arctan (x))$ .

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Step 2.4.1.1

Draw a triangle in the plane with vertices $(1, x)$ , $(1, 0)$ , and the origin. Then $\arctan (x)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(1, x)$ . Therefore, $\cos (\arctan (x))$ is $\frac{1}{\sqrt{1 + x^{2}}}$ .

$6 y = \frac{1}{\sqrt{1 + x^{2}}}$

Step 2.4.1.2

Multiply $\frac{1}{\sqrt{1 + x^{2}}}$ by $\frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}$ .

$6 y = \frac{1}{\sqrt{1 + x^{2}}} \cdot \frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}$

Step 2.4.1.3

Combine and simplify the denominator.

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Step 2.4.1.3.1

Multiply $\frac{1}{\sqrt{1 + x^{2}}}$ by $\frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}$ .

$6 y = \frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}} \sqrt{1 + x^{2}}}$

Step 2.4.1.3.2

Raise $\sqrt{1 + x^{2}}$ to the power of $1$ .

$6 y = \frac{\sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{1} \sqrt{1 + x^{2}}}$

Step 2.4.1.3.3

Raise $\sqrt{1 + x^{2}}$ to the power of $1$ .

$6 y = \frac{\sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{1} {\sqrt{1 + x^{2}}}^{1}}$

Step 2.4.1.3.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$6 y = \frac{\sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{1 + 1}}$

Step 2.4.1.3.5

Add $1$ and $1$ .

$6 y = \frac{\sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{2}}$

Step 2.4.1.3.6

Rewrite ${\sqrt{1 + x^{2}}}^{2}$ as $1 + x^{2}$ .

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Step 2.4.1.3.6.1

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

$6 y = \frac{\sqrt{1 + x^{2}}}{{({(1 + x^{2})}^{\frac{1}{2}})}^{2}}$

Step 2.4.1.3.6.2

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

$6 y = \frac{\sqrt{1 + x^{2}}}{{(1 + x^{2})}^{\frac{1}{2} \cdot 2}}$

Step 2.4.1.3.6.3

Combine $\frac{1}{2}$ and $2$ .

$6 y = \frac{\sqrt{1 + x^{2}}}{{(1 + x^{2})}^{\frac{2}{2}}}$

Step 2.4.1.3.6.4

Cancel the common factor of $2$ .

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Step 2.4.1.3.6.4.1

Cancel the common factor.

$6 y = \frac{\sqrt{1 + x^{2}}}{{(1 + x^{2})}^{\frac{2}{2}}}$

Step 2.4.1.3.6.4.2

Rewrite the expression.

$6 y = \frac{\sqrt{1 + x^{2}}}{{(1 + x^{2})}^{1}}$

Step 2.4.1.3.6.5

Simplify.

$6 y = \frac{\sqrt{1 + x^{2}}}{1 + x^{2}}$

Step 2.5

Divide each term in $6 y = \frac{\sqrt{1 + x^{2}}}{1 + x^{2}}$ by $6$ and simplify.

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Step 2.5.1

Divide each term in $6 y = \frac{\sqrt{1 + x^{2}}}{1 + x^{2}}$ by $6$ .

$\frac{6 y}{6} = \frac{\frac{\sqrt{1 + x^{2}}}{1 + x^{2}}}{6}$

Step 2.5.2

Simplify the left side.

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Step 2.5.2.1

Cancel the common factor of $6$ .

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Step 2.5.2.1.1

Cancel the common factor.

$\frac{6 y}{6} = \frac{\frac{\sqrt{1 + x^{2}}}{1 + x^{2}}}{6}$

Step 2.5.2.1.2

Divide $y$ by $1$ .

$y = \frac{\frac{\sqrt{1 + x^{2}}}{1 + x^{2}}}{6}$

Step 2.5.3

Simplify the right side.

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Step 2.5.3.1

Multiply the numerator by the reciprocal of the denominator.

$y = \frac{\sqrt{1 + x^{2}}}{1 + x^{2}} \cdot \frac{1}{6}$

Step 2.5.3.2

Multiply $\frac{\sqrt{1 + x^{2}}}{1 + x^{2}}$ by $\frac{1}{6}$ .

$y = \frac{\sqrt{1 + x^{2}}}{(1 + x^{2}) \cdot 6}$

Step 2.5.3.3

Move $6$ to the left of $1 + x^{2}$ .

$y = \frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}$

Step 3

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

$f^{- 1} (x) = \frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}$

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}$ is the inverse of $f (x) = \tan (\arccos (6 x))$ .

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Step 4.1

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Step 4.2

Evaluate $f^{- 1} (f (x))$ .

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Step 4.2.1

Set up the composite result function.

$f^{- 1} (f (x))$

Step 4.2.2

Evaluate $f^{- 1} (\tan (\arccos (6 x)))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\sqrt{1 + \tan^{2} (\arccos (6 x))}}{6 (1 + \tan^{2} (\arccos (6 x)))}$

Step 4.2.3

Rearrange terms.

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\sqrt{\tan^{2} (\arccos (6 x)) + 1}}{6 (1 + \tan^{2} (\arccos (6 x)))}$

Step 4.2.4

Apply pythagorean identity.

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\sqrt{\sec^{2} (\arccos (6 x))}}{6 (1 + \tan^{2} (\arccos (6 x)))}$

Step 4.2.5

Rearrange terms.

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\sqrt{\sec^{2} (\arccos (6 x))}}{6 (\tan^{2} (\arccos (6 x)) + 1)}$

Step 4.2.6

Apply pythagorean identity.

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\sqrt{\sec^{2} (\arccos (6 x))}}{6 \sec^{2} (\arccos (6 x))}$

Step 4.2.7

Simplify the numerator.

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Step 4.2.7.1

Pull terms out from under the radical, assuming positive real numbers.

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\sec (\arccos (6 x))}{6 \sec^{2} (\arccos (6 x))}$

Step 4.2.7.2

Draw a triangle in the plane with vertices $(6 x, \sqrt{1^{2} - {(6 x)}^{2}})$ , $(6 x, 0)$ , and the origin. Then $\arccos (6 x)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(6 x, \sqrt{1^{2} - {(6 x)}^{2}})$ . Therefore, $\sec (\arccos (6 x))$ is $\frac{1}{6 x}$ .

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\frac{1}{6 x}}{6 \sec^{2} (\arccos (6 x))}$

Step 4.2.8

Simplify the denominator.

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Step 4.2.8.1

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\frac{1}{6 x}}{6 {(\frac{1}{6 x})}^{2}}$

Step 4.2.8.2

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 4.2.8.2.1

Apply the product rule to $\frac{1}{6 x}$ .

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\frac{1}{6 x}}{6 (\frac{1^{2}}{{(6 x)}^{2}})}$

Step 4.2.8.2.2

Apply the product rule to $6 x$ .

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\frac{1}{6 x}}{6 (\frac{1^{2}}{6^{2} x^{2}})}$

Step 4.2.8.3

One to any power is one.

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\frac{1}{6 x}}{6 (\frac{1}{6^{2} x^{2}})}$

Step 4.2.8.4

Raise $6$ to the power of $2$ .

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\frac{1}{6 x}}{6 (\frac{1}{36 x^{2}})}$

Step 4.2.9

Simplify terms.

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Step 4.2.9.1

Combine $6$ and $\frac{1}{36 x^{2}}$ .

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\frac{1}{6 x}}{\frac{6}{36 x^{2}}}$

Step 4.2.9.2

Reduce the expression $\frac{6}{36 x^{2}}$ by cancelling the common factors.

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Step 4.2.9.2.1

Factor $6$ out of $6$ .

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\frac{1}{6 x}}{\frac{6 (1)}{36 x^{2}}}$

Step 4.2.9.2.2

Factor $6$ out of $36 x^{2}$ .

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\frac{1}{6 x}}{\frac{6 (1)}{6 (6 x^{2})}}$

Step 4.2.9.2.3

Cancel the common factor.

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\frac{1}{6 x}}{\frac{6 \cdot 1}{6 (6 x^{2})}}$

Step 4.2.9.2.4

Rewrite the expression.

$f^{- 1} (\tan (\arccos (6 x))) = \frac{\frac{1}{6 x}}{\frac{1}{6 x^{2}}}$

Step 4.2.10

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\tan (\arccos (6 x))) = \frac{1}{6 x} \cdot (6 x^{2})$

Step 4.2.11

Rewrite using the commutative property of multiplication.

$f^{- 1} (\tan (\arccos (6 x))) = 6 (\frac{1}{6 x}) \cdot x^{2}$

Step 4.2.12

Cancel the common factor of $6$ .

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Step 4.2.12.1

Factor $6$ out of $6 x$ .

$f^{- 1} (\tan (\arccos (6 x))) = 6 (\frac{1}{6 (x)}) \cdot x^{2}$

Step 4.2.12.2

Cancel the common factor.

$f^{- 1} (\tan (\arccos (6 x))) = 6 (\frac{1}{6 x}) \cdot x^{2}$

Step 4.2.12.3

Rewrite the expression.

$f^{- 1} (\tan (\arccos (6 x))) = \frac{1}{x} \cdot x^{2}$

Step 4.2.13

Cancel the common factor of $x$ .

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Step 4.2.13.1

Factor $x$ out of $x^{2}$ .

$f^{- 1} (\tan (\arccos (6 x))) = \frac{1}{x} \cdot (x \cdot x)$

Step 4.2.13.2

Cancel the common factor.

$f^{- 1} (\tan (\arccos (6 x))) = \frac{1}{x} \cdot (x \cdot x)$

Step 4.2.13.3

Rewrite the expression.

$f^{- 1} (\tan (\arccos (6 x))) = x$

Step 4.3

Evaluate $f (f^{- 1} (x))$ .

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Step 4.3.1

Set up the composite result function.

$f (f^{- 1} (x))$

Step 4.3.2

Evaluate $f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \tan (\arccos (6 (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})})))$

Step 4.3.3

Draw a triangle in the plane with vertices $(6 (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}), \sqrt{1^{2} - {(6 (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}))}^{2}})$ , $(6 (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}), 0)$ , and the origin. Then $\arccos (6 (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}))$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(6 (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}), \sqrt{1^{2} - {(6 (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}))}^{2}})$ . Therefore, $\tan (\arccos (6 (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})})))$ is $\frac{\sqrt{1 - {(\frac{6 \sqrt{1 + x^{2}}}{6 (1 + x^{2})})}^{2}}}{\frac{6 \sqrt{1 + x^{2}}}{6 (1 + x^{2})}}$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{1 - {(\frac{6 \sqrt{1 + x^{2}}}{6 (1 + x^{2})})}^{2}}}{\frac{6 \sqrt{1 + x^{2}}}{6 (1 + x^{2})}}$

Step 4.3.4

Cancel the common factor of $6$ .

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Step 4.3.4.1

Cancel the common factor.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{1 - {(\frac{6 \sqrt{1 + x^{2}}}{6 (1 + x^{2})})}^{2}}}{\frac{6 \sqrt{1 + x^{2}}}{6 (1 + x^{2})}}$

Step 4.3.4.2

Rewrite the expression.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{1 - {(\frac{\sqrt{1 + x^{2}}}{1 + x^{2}})}^{2}}}{\frac{6 \sqrt{1 + x^{2}}}{6 (1 + x^{2})}}$

Step 4.3.5

Cancel the common factor of $6$ .

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Step 4.3.5.1

Cancel the common factor.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{1 - {(\frac{\sqrt{1 + x^{2}}}{1 + x^{2}})}^{2}}}{\frac{6 \sqrt{1 + x^{2}}}{6 (1 + x^{2})}}$

Step 4.3.5.2

Rewrite the expression.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{1 - {(\frac{\sqrt{1 + x^{2}}}{1 + x^{2}})}^{2}}}{\frac{\sqrt{1 + x^{2}}}{1 + x^{2}}}$

Step 4.3.6

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \sqrt{1 - {(\frac{\sqrt{1 + x^{2}}}{1 + x^{2}})}^{2}} (\frac{1 + x^{2}}{\sqrt{1 + x^{2}}})$

Step 4.3.7

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

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \sqrt{1^{2} - {(\frac{\sqrt{1 + x^{2}}}{1 + x^{2}})}^{2}} (\frac{1 + x^{2}}{\sqrt{1 + x^{2}}})$

Step 4.3.8

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 $b = \frac{\sqrt{1 + x^{2}}}{1 + x^{2}}$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \sqrt{(1 + \frac{\sqrt{1 + x^{2}}}{1 + x^{2}}) (1 - \frac{\sqrt{1 + x^{2}}}{1 + x^{2}})} (\frac{1 + x^{2}}{\sqrt{1 + x^{2}}})$

Step 4.3.9

Simplify.

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Step 4.3.9.1

Write $1$ as a fraction with a common denominator.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \sqrt{(\frac{1 + x^{2}}{1 + x^{2}} + \frac{\sqrt{1 + x^{2}}}{1 + x^{2}}) (1 - \frac{\sqrt{1 + x^{2}}}{1 + x^{2}})} (\frac{1 + x^{2}}{\sqrt{1 + x^{2}}})$

Step 4.3.9.2

Combine the numerators over the common denominator.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \sqrt{\frac{1 + x^{2} + \sqrt{1 + x^{2}}}{1 + x^{2}} \cdot (1 - \frac{\sqrt{1 + x^{2}}}{1 + x^{2}})} (\frac{1 + x^{2}}{\sqrt{1 + x^{2}}})$

Step 4.3.9.3

Write $1$ as a fraction with a common denominator.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \sqrt{\frac{1 + x^{2} + \sqrt{1 + x^{2}}}{1 + x^{2}} \cdot (\frac{1 + x^{2}}{1 + x^{2}} - \frac{\sqrt{1 + x^{2}}}{1 + x^{2}})} (\frac{1 + x^{2}}{\sqrt{1 + x^{2}}})$

Step 4.3.9.4

Combine the numerators over the common denominator.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \sqrt{\frac{1 + x^{2} + \sqrt{1 + x^{2}}}{1 + x^{2}} \cdot \frac{1 + x^{2} - \sqrt{1 + x^{2}}}{1 + x^{2}}} (\frac{1 + x^{2}}{\sqrt{1 + x^{2}}})$

Step 4.3.10

Multiply $\frac{1 + x^{2} + \sqrt{1 + x^{2}}}{1 + x^{2}}$ by $\frac{1 + x^{2} - \sqrt{1 + x^{2}}}{1 + x^{2}}$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \sqrt{\frac{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})}{(1 + x^{2}) (1 + x^{2})}} (\frac{1 + x^{2}}{\sqrt{1 + x^{2}}})$

Step 4.3.11

Simplify the denominator.

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Step 4.3.11.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \sqrt{\frac{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})}{{(1 + x^{2})}^{1 + 1}}} (\frac{1 + x^{2}}{\sqrt{1 + x^{2}}})$

Step 4.3.11.2

Add $1$ and $1$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \sqrt{\frac{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})}{{(1 + x^{2})}^{2}}} (\frac{1 + x^{2}}{\sqrt{1 + x^{2}}})$

Step 4.3.12

Rewrite $\frac{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})}{{(1 + x^{2})}^{2}}$ as ${(\frac{1}{1 + x^{2}})}^{2} ((1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}}))$ .

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Step 4.3.12.1

Factor the perfect power $1^{2}$ out of $(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \sqrt{\frac{1^{2} ((1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}}))}{{(1 + x^{2})}^{2}}} (\frac{1 + x^{2}}{\sqrt{1 + x^{2}}})$

Step 4.3.12.2

Factor the perfect power ${(1 + x^{2})}^{2}$ out of ${(1 + x^{2})}^{2}$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \sqrt{\frac{1^{2} ((1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}}))}{{(1 + x^{2})}^{2} \cdot 1}} (\frac{1 + x^{2}}{\sqrt{1 + x^{2}}})$

Step 4.3.12.3

Rearrange the fraction $\frac{1^{2} ((1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}}))}{{(1 + x^{2})}^{2} \cdot 1}$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \sqrt{{(\frac{1}{1 + x^{2}})}^{2} ((1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}}))} (\frac{1 + x^{2}}{\sqrt{1 + x^{2}}})$

Step 4.3.13

Pull terms out from under the radical.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{1}{1 + x^{2}} \cdot (\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})} (\frac{1 + x^{2}}{\sqrt{1 + x^{2}}}))$

Step 4.3.14

Combine $\frac{1}{1 + x^{2}}$ and $\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})}$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})}}{1 + x^{2}} \cdot \frac{1 + x^{2}}{\sqrt{1 + x^{2}}}$

Step 4.3.15

Cancel the common factor of $1 + x^{2}$ .

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Step 4.3.15.1

Cancel the common factor.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})}}{1 + x^{2}} \cdot \frac{1 + x^{2}}{\sqrt{1 + x^{2}}}$

Step 4.3.15.2

Rewrite the expression.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})} (\frac{1}{\sqrt{1 + x^{2}}})$

Step 4.3.16

Combine $\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})}$ and $\frac{1}{\sqrt{1 + x^{2}}}$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})}}{\sqrt{1 + x^{2}}}$

Step 4.3.17

Multiply $\frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})}}{\sqrt{1 + x^{2}}}$ by $\frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})}}{\sqrt{1 + x^{2}}} \cdot \frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}$

Step 4.3.18

Combine and simplify the denominator.

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Step 4.3.18.1

Multiply $\frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})}}{\sqrt{1 + x^{2}}}$ by $\frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})} \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}} \sqrt{1 + x^{2}}}$

Step 4.3.18.2

Raise $\sqrt{1 + x^{2}}$ to the power of $1$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})} \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}} \sqrt{1 + x^{2}}}$

Step 4.3.18.3

Raise $\sqrt{1 + x^{2}}$ to the power of $1$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})} \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}} \sqrt{1 + x^{2}}}$

Step 4.3.18.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})} \sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{1 + 1}}$

Step 4.3.18.5

Add $1$ and $1$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})} \sqrt{1 + x^{2}}}{{\sqrt{1 + x^{2}}}^{2}}$

Step 4.3.18.6

Rewrite ${\sqrt{1 + x^{2}}}^{2}$ as $1 + x^{2}$ .

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Step 4.3.18.6.1

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

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})} \sqrt{1 + x^{2}}}{{({(1 + x^{2})}^{\frac{1}{2}})}^{2}}$

Step 4.3.18.6.2

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

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})} \sqrt{1 + x^{2}}}{{(1 + x^{2})}^{\frac{1}{2} \cdot 2}}$

Step 4.3.18.6.3

Combine $\frac{1}{2}$ and $2$ .

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})} \sqrt{1 + x^{2}}}{{(1 + x^{2})}^{\frac{2}{2}}}$

Step 4.3.18.6.4

Cancel the common factor of $2$ .

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Step 4.3.18.6.4.1

Cancel the common factor.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})} \sqrt{1 + x^{2}}}{{(1 + x^{2})}^{\frac{2}{2}}}$

Step 4.3.18.6.4.2

Rewrite the expression.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})} \sqrt{1 + x^{2}}}{1 + x^{2}}$

Step 4.3.18.6.5

Simplify.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}})} \sqrt{1 + x^{2}}}{1 + x^{2}}$

Step 4.3.19

Combine using the product rule for radicals.

$f (\frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}) = \frac{\sqrt{(1 + x^{2} + \sqrt{1 + x^{2}}) (1 + x^{2} - \sqrt{1 + x^{2}}) (1 + x^{2})}}{1 + x^{2}}$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}$ is the inverse of $f (x) = \tan (\arccos (6 x))$ .

$f^{- 1} (x) = \frac{\sqrt{1 + x^{2}}}{6 (1 + x^{2})}$