Double Angle Formulas: Finding $\tan 2\theta$

Question

I am trying to find $\tan 2\theta$ where $sin \theta = \frac{5}{13}$ and $\theta$ is in Quadrant One.

According to my textbook, $\tan 2\theta = \frac{120}{119}$, but I get $\frac{-10}{13}$ instead.

The Identity I am using:

$$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^{2}\theta}$$

My Process:

Since $y = 5\;$ and $r = 13,\; x = 12.$

Apply Tangent Double Angle Formula: $$\frac{2(\frac{5}{12})}{1 - (\frac{5}{12})^2}$$

$$\frac{\frac{10}{12}}{1 - \frac{25}{12}} \cdot \frac{12}{12}$$

$$\frac{10}{12-25}$$

$$\frac{-10}{13}$$

What am I doing wrong?

Answer 1

Alternatively:

$\sin\theta = 5/13$, implies $\cos\theta = \sqrt{1-(5/13)^2} = 12/13$.

$\sin 2\theta = 2\sin \theta \cos \theta = 120/169$.

$\cos 2\theta = 2 \cos^2 \theta-1 = 1 - 2 \sin^2 \theta = 119/169$.

This gives $\tan 2\theta = \sin 2\theta/ \cos 2 \theta = 120/119$.

Answer 2

I made a mistake solving the problem. $(\frac{5}{12})^2 \neq \frac{25}{12}$.

Actually, $(\frac{5}{12})^2 = \frac{25}{144}$.

Taking that into account:

$$\frac{\frac{10}{12}}{1 - \frac{25}{144}} \cdot \frac{12}{12}$$

$$\frac{10}{12-\frac{300}{144}}$$

$$\frac{10}{12 - \frac{25}{12}} \cdot \frac{12}{12}$$

$$\frac{120}{144-25}$$

$$\frac{120}{119}$$

Therefore, $\tan 2\theta = \frac{120}{119}$