1. **Problem statement:** Given $\sin A = \frac{6}{7}$ and $A$ is in quadrant II, find $\sin 2A$, $\cos 2A$, and $\tan 2A$. 2. **Recall formulas:** - Double angle formulas: $$\sin 2A = 2 \sin A \cos A$$ $$\cos 2A = \cos^2 A - \sin^2 A$$ $$\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}$$ 3. **Find $\cos A$:** Since $\sin^2 A + \cos^2 A = 1$, $$\cos A = \pm \sqrt{1 - \sin^2 A} = \pm \sqrt{1 - \left(\frac{6}{7}\right)^2} = \pm \sqrt{1 - \frac{36}{49}} = \pm \sqrt{\frac{13}{49}} = \pm \frac{\sqrt{13}}{7}$$ Because $A$ is in quadrant II, where cosine is negative, $$\cos A = - \frac{\sqrt{13}}{7}$$ 4. **Calculate $\sin 2A$:** $$\sin 2A = 2 \sin A \cos A = 2 \times \frac{6}{7} \times \left(- \frac{\sqrt{13}}{7}\right) = - \frac{12 \sqrt{13}}{49}$$ 5. **Calculate $\cos 2A$:** $$\cos 2A = \cos^2 A - \sin^2 A = \left(- \frac{\sqrt{13}}{7}\right)^2 - \left(\frac{6}{7}\right)^2 = \frac{13}{49} - \frac{36}{49} = - \frac{23}{49}$$ 6. **Calculate $\tan A$:** $$\tan A = \frac{\sin A}{\cos A} = \frac{\frac{6}{7}}{- \frac{\sqrt{13}}{7}} = - \frac{6}{\sqrt{13}} = - \frac{6 \sqrt{13}}{13}$$ 7. **Calculate $\tan 2A$:** $$\tan 2A = \frac{2 \tan A}{1 - \tan^2 A} = \frac{2 \times \left(- \frac{6 \sqrt{13}}{13}\right)}{1 - \left(- \frac{6 \sqrt{13}}{13}\right)^2} = \frac{- \frac{12 \sqrt{13}}{13}}{1 - \frac{36 \times 13}{169}} = \frac{- \frac{12 \sqrt{13}}{13}}{1 - \frac{468}{169}} = \frac{- \frac{12 \sqrt{13}}{13}}{\frac{169 - 468}{169}} = \frac{- \frac{12 \sqrt{13}}{13}}{- \frac{299}{169}}$$ 8. **Simplify $\tan 2A$:** $$\tan 2A = - \frac{12 \sqrt{13}}{13} \times \frac{169}{-299} = \frac{12 \sqrt{13} \times 169}{13 \times 299} = \frac{12 \times 169 \sqrt{13}}{3887}$$ Simplify numerator and denominator: $$12 \times 169 = 2028$$ So, $$\tan 2A = \frac{2028 \sqrt{13}}{3887}$$ **Final answers:** $$\sin 2A = - \frac{12 \sqrt{13}}{49}$$ $$\cos 2A = - \frac{23}{49}$$ $$\tan 2A = \frac{2028 \sqrt{13}}{3887}$$