1. **Problem:** Find the value of $\sec(\arctan(\frac{2}{3}))$. 2. **Formula and rules:** Recall that $\sec(\theta) = \frac{1}{\cos(\theta)}$ and $\arctan(x)$ gives an angle $\theta$ such that $\tan(\theta) = x$. 3. **Step-by-step solution:** - Let $\theta = \arctan(\frac{2}{3})$, so $\tan(\theta) = \frac{2}{3}$. - Consider a right triangle where the opposite side to $\theta$ is 2 and the adjacent side is 3. - The hypotenuse $h$ is given by the Pythagorean theorem: $$h = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}.$$ - Then, $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{3}{\sqrt{13}}$. - Therefore, $$\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{\frac{3}{\sqrt{13}}} = \frac{\sqrt{13}}{3}.$$ 4. **Matching with options:** The options given are fractions with denominators $\sqrt{14}$, $\sqrt{13}$, $\sqrt{15}$, etc. Our answer $\frac{\sqrt{13}}{3}$ is not exactly any of the options. - Option A: $\frac{3}{\sqrt{14}}$ - Option B: $\frac{2}{\sqrt{13}}$ - Option C: None of these - Option D: $\frac{1}{\sqrt{15}}$ Since $\frac{\sqrt{13}}{3} \neq \frac{2}{\sqrt{13}}$ and none of the other options match, the correct choice is C. **Final answer:** C. None of these