1. **Problem Statement:** Given that $\sec \theta = a + \frac{1}{4a}$, find the value of $\sec \theta + \tan \theta$. 2. **Recall the identity:** $$\sec^2 \theta - \tan^2 \theta = 1$$ This implies: $$\tan^2 \theta = \sec^2 \theta - 1$$ 3. **Calculate $\sec^2 \theta$:** $$\sec \theta = a + \frac{1}{4a}$$ Square both sides: $$\sec^2 \theta = \left(a + \frac{1}{4a}\right)^2 = a^2 + 2 \cdot a \cdot \frac{1}{4a} + \left(\frac{1}{4a}\right)^2 = a^2 + \frac{1}{2} + \frac{1}{16a^2}$$ 4. **Calculate $\tan^2 \theta$:** $$\tan^2 \theta = \sec^2 \theta - 1 = \left(a^2 + \frac{1}{2} + \frac{1}{16a^2}\right) - 1 = a^2 - \frac{1}{2} + \frac{1}{16a^2}$$ 5. **Calculate $\tan \theta$:** Since $\tan \theta$ can be positive or negative, but we assume positive for this problem, $$\tan \theta = \sqrt{a^2 - \frac{1}{2} + \frac{1}{16a^2}}$$ 6. **Calculate $\sec \theta + \tan \theta$:** We want to simplify: $$a + \frac{1}{4a} + \sqrt{a^2 - \frac{1}{2} + \frac{1}{16a^2}}$$ 7. **Simplify the expression under the square root:** Rewrite the radicand as: $$a^2 - \frac{1}{2} + \frac{1}{16a^2} = \left(a - \frac{1}{4a}\right)^2$$ Because: $$\left(a - \frac{1}{4a}\right)^2 = a^2 - 2 \cdot a \cdot \frac{1}{4a} + \left(\frac{1}{4a}\right)^2 = a^2 - \frac{1}{2} + \frac{1}{16a^2}$$ 8. **Therefore:** $$\tan \theta = \left|a - \frac{1}{4a}\right|$$ Assuming $a > 0$, we take positive value: $$\tan \theta = a - \frac{1}{4a}$$ 9. **Sum $\sec \theta + \tan \theta$:** $$\left(a + \frac{1}{4a}\right) + \left(a - \frac{1}{4a}\right) = a + \frac{1}{4a} + a - \frac{1}{4a} = 2a$$ **Final answer:** $$\boxed{2a}$$