(3) If \( \cos _{\substack{\text { dousbe } \\ \text { angle. }}}^{\cos 2 A}=\frac{12}{13} \), find 2 possible values of tan \( A \) (withoust using a calculutor)

To solve the problem, we start from the identity: \[ \cos 2A = 2\cos^2 A - 1 \] Since it's given that \(\cos 2A = \frac{12}{13}\), we can set up the equation: \[ 2\cos^2 A - 1 = \frac{12}{13} \] Adding 1 to both sides gives: \[ 2\cos^2 A = \frac{12}{13} + 1 = \frac{12}{13} + \frac{13}{13} = \frac{25}{13} \] Now, divide both sides by 2: \[ \cos^2 A = \frac{25}{26} \] Taking the square root of both sides, we find: \[ \cos A = \pm \sqrt{\frac{25}{26}} = \pm \frac{5}{\sqrt{26}} = \pm \frac{5\sqrt{26}}{26} \] Next, we will use the identity \(\tan A = \frac{\sin A}{\cos A}\). We know that \(\sin^2 A + \cos^2 A = 1\). So, \[ \sin^2 A = 1 - \cos^2 A = 1 - \frac{25}{26} = \frac{1}{26} \] Thus, \[ \sin A = \pm \sqrt{\frac{1}{26}} = \pm \frac{1}{\sqrt{26}} = \pm \frac{\sqrt{26}}{26} \] Now we can find the possible values of \(\tan A\): \[ \tan A = \frac{\sin A}{\cos A} = \frac{\pm \frac{1}{\sqrt{26}}}{\pm \frac{5}{\sqrt{26}}} = \frac{1}{5} \text{ or } -\frac{1}{5} \] Hence, the two possible values of \(\tan A\) are \(\frac{1}{5}\) and \(-\frac{1}{5}\).