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If $\cos \theta=\frac{4}{\sqrt{26}}$ and angle $\theta$ is in Quadrant I, what is the exact value of $\tan 2 \theta$ in simplest radical form?

Studdy Solution
Use the double angle identity for tangent:
$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$
First, find $\tan \theta$:
$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{\frac{5}{13}}}{\frac{4}{\sqrt{26}}}$
Simplify $\tan \theta$:
$\tan \theta = \frac{\sqrt{5/13} \cdot \sqrt{26}}{4} = \frac{\sqrt{5} \cdot \sqrt{2}}{4} = \frac{\sqrt{10}}{4}$
Now, substitute $\tan \theta$ into the double angle formula:
$\tan 2\theta = \frac{2 \cdot \frac{\sqrt{10}}{4}}{1 - \left(\frac{\sqrt{10}}{4}\right)^2}$
Simplify the expression:
$\tan 2\theta = \frac{\frac{\sqrt{10}}{2}}{1 - \frac{10}{16}}$
$\tan 2\theta = \frac{\frac{\sqrt{10}}{2}}{\frac{6}{16}}$
$\tan 2\theta = \frac{\sqrt{10}}{2} \cdot \frac{16}{6}$
$\tan 2\theta = \frac{8\sqrt{10}}{6}$
Simplify further:
$\tan 2\theta = \frac{4\sqrt{10}}{3}$
The exact value of $\tan 2\theta$ is:
$\boxed{\frac{4\sqrt{10}}{3}}$