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If cos θ= 12 /13 and θ is located in the Quadrant I, find sin (2 θ ), cos(2 θ ), tan(2 θ )

$$\cos 2 \theta = \dfrac{119}{169}\\\\\sin2 \theta = \dfrac{120}{169}\\\\\tan 2\theta = \dfrac{120}{119}$$
$$\text{Given that,} \cos \theta = \dfrac{12}{13}\\ \\\text{Now,}\\\\\cos 2 \theta = 2 \cos^2 \theta - 1\\\\\\~~~~~~~~~=2 \left( \dfrac{12}{13} \right)^2 - 1\\\\\\~~~~~~~~~=2 \left( \dfrac{144}{169} \right) - 1\\\\\\~~~~~~~~~=\dfrac{288}{169}-1\\\\\\~~~~~~~~~=\dfrac{119}{169}$$
$$\sin 2\theta = 2 \sin \theta \cos \theta\\\\\\~~~~~~~~=2\sqrt{1 -\cos^2 \theta} \cdot \cos \theta~~~~~~~~~~~;[\text{In quadrant I, all ratios are positive.}]\\\\\\~~~~~~~~=2 \sqrt{1- \left( \dfrac{12}{13} \right)^2} \cdot \left(\dfrac{12}{13} \right)\\\\\\~~~~~~~~=\left( \dfrac{24}{13} \right) \sqrt{1- \dfrac{144}{169}}\\\\\\~~~~~~~~=\dfrac{24}{13}\sqrt{\dfrac{25}{169}}\\\\\\~~~~~~~=\dfrac{24}{13} \times \dfrac{5}{13}\\\\\\~~~~~~=\dfrac{120}{169}$$
$$\tan 2 \theta = \dfrac{\sin 2\theta}{\cos 2\theta}\\\\\\~~~~~~~~~=\dfrac{\tfrac{120}{169}}{ \tfrac{119}{169}}\\\\\\~~~~~~~~~=\dfrac{120}{169} \times \dfrac{169}{119}\\\\\\~~~~~~~~~=\dfrac{120}{119}$$