Solving the identity, $$ an (A)= $$ $$ \pm \frac{7}{9} $$ $$ \pm \frac{\sqrt{7}}{9} $$

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Solving the identity, $$ 	an (A)= $$ $$ \pm \frac{7}{9} $$ $$ \pm \frac{\sqrt{7}}{9} $$

UpStudy AI · Accepted Answer

To solve the identity $$ \tan(A) = \pm \frac{7}{9} $$ and $$ \tan(A) = \pm \frac{\sqrt{7}}{9} $$, we will analyze each case separately.

### Case 1: $$ \tan(A) = \pm \frac{7}{9} $$

1. **Understanding the Tangent Function**: The tangent function is defined as the ratio of the opposite side to the adjacent side in a right triangle. Therefore, if $$ \tan(A) = \frac{7}{9} $$, we can visualize a right triangle where the opposite side is 7 and the adjacent side is 9.

2. **Finding the Angle**: To find the angle $$ A $$, we can use the arctangent function:
   $$
   A = \tan^{-1}\left(\frac{7}{9}\right)
   $$
   This will give us one solution for $$ A $$. Since tangent is periodic with a period of $$ \pi $$, the general solutions will be:
   $$
   A = \tan^{-1}\left(\frac{7}{9}\right) + n\pi \quad \text{for } n \in \mathbb{Z}
   $$
   Similarly, for $$ \tan(A) = -\frac{7}{9} $$:
   $$
   A = \tan^{-1}\left(-\frac{7}{9}\right) + n\pi \quad \text{for } n \in \mathbb{Z}
   $$

### Case 2: $$ \tan(A) = \pm \frac{\sqrt{7}}{9} $$

1. **Understanding the Tangent Function**: Similarly, if $$ \tan(A) = \frac{\sqrt{7}}{9} $$, we can visualize a right triangle where the opposite side is $$ \sqrt{7} $$ and the adjacent side is 9.

2. **Finding the Angle**: To find the angle $$ A $$, we can use the arctangent function:
   $$
   A = \tan^{-1}\left(\frac{\sqrt{7}}{9}\right)
   $$
   The general solutions will be:
   $$
   A = \tan^{-1}\left(\frac{\sqrt{7}}{9}\right) + n\pi \quad \text{for } n \in \mathbb{Z}
   $$
   For $$ \tan(A) = -\frac{\sqrt{7}}{9} $$:
   $$
   A = \tan^{-1}\left(-\frac{\sqrt{7}}{9}\right) + n\pi \quad \text{for } n \in \mathbb{Z}
   $$

### Summary of Solutions

- For $$ \tan(A) = \frac{7}{9} $$:
  $$
  A = \tan^{-1}\left(\frac{7}{9}\right) + n\pi
  $$
- For $$ \tan(A) = -\frac{7}{9} $$:
  $$
  A = \tan^{-1}\left(-\frac{7}{9}\right) + n\pi
  $$
- For $$ \tan(A) = \frac{\sqrt{7}}{9} $$:
  $$
  A = \tan^{-1}\left(\frac{\sqrt{7}}{9}\right) + n\pi
  $$
- For $$ \tan(A) = -\frac{\sqrt{7}}{9} $$:
  $$
  A = \tan^{-1}\left(-\frac{\sqrt{7}}{9}\right) + n\pi
  $$

These equations provide the angles $$ A $$ for each case, where $$ n $$ is any integer.
The solutions for $$ \tan(A) $$ are: - $$ A = \tan^{-1}\left(\frac{7}{9}\right) + n\pi $$ - $$ A = \tan^{-1}\left(-\frac{7}{9}\right) + n\pi $$ - $$ A = \tan^{-1}\left(\frac{\sqrt{7}}{9}\right) + n\pi $$ - $$ A = \tan^{-1}\left(-\frac{\sqrt{7}}{9}\right) + n\pi $$ where $$ n $$ is any integer.

Solving the identity, \( \tan (A)= \) \( \pm \frac{7}{9} \) \( \pm \frac{\sqrt{7}}{9} \)

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