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

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. - The values \( \frac{7}{9} \) and \( -\frac{7}{9} \) indicate that the angle \( A \) can be in the first or third quadrant (for positive) and the second or fourth quadrant (for negative). 2. **Finding the Angle**: - We can find the reference angle using the arctangent function: \[ A = \tan^{-1}\left(\frac{7}{9}\right) \] - The angles in the respective quadrants will be: - First Quadrant: \( A = \tan^{-1}\left(\frac{7}{9}\right) \) - Third Quadrant: \( A = \tan^{-1}\left(\frac{7}{9}\right) + \pi \) - Second Quadrant: \( A = \pi - \tan^{-1}\left(\frac{7}{9}\right) \) - Fourth Quadrant: \( A = 2\pi - \tan^{-1}\left(\frac{7}{9}\right) \) ### Case 2: \( \tan(A) = \pm \frac{\sqrt{7}}{9} \) 1. **Understanding the Tangent Function**: - Similar to the previous case, the values \( \frac{\sqrt{7}}{9} \) and \( -\frac{\sqrt{7}}{9} \) indicate that the angle \( A \) can be in the first or third quadrant (for positive) and the second or fourth quadrant (for negative). 2. **Finding the Angle**: - We can find the reference angle using the arctangent function: \[ A = \tan^{-1}\left(\frac{\sqrt{7}}{9}\right) \] - The angles in the respective quadrants will be: - First Quadrant: \( A = \tan^{-1}\left(\frac{\sqrt{7}}{9}\right) \) - Third Quadrant: \( A = \tan^{-1}\left(\frac{\sqrt{7}}{9}\right) + \pi \) - Second Quadrant: \( A = \pi - \tan^{-1}\left(\frac{\sqrt{7}}{9}\right) \) - Fourth Quadrant: \( A = 2\pi - \tan^{-1}\left(\frac{\sqrt{7}}{9}\right) \) ### Summary of Solutions - For \( \tan(A) = \pm \frac{7}{9} \): - \( A = \tan^{-1}\left(\frac{7}{9}\right) \) - \( A = \tan^{-1}\left(\frac{7}{9}\right) + \pi \) - \( A = \pi - \tan^{-1}\left(\frac{7}{9}\right) \) - \( A = 2\pi - \tan^{-1}\left(\frac{7}{9}\right) \) - For \( \tan(A) = \pm \frac{\sqrt{7}}{9} \): - \( A = \tan^{-1}\left(\frac{\sqrt{7}}{9}\right) \) - \( A = \tan^{-1}\left(\frac{\sqrt{7}}{9}\right) + \pi \) - \( A = \pi - \tan^{-1}\left(\frac{\sqrt{7}}{9}\right) \) - \( A = 2\pi - \tan^{-1}\left(\frac{\sqrt{7}}{9}\right) \) These angles represent all possible solutions for the given identities.