1. **State the problem:** Given the equation $7\cos\theta - \sin\theta = 5$ and the condition $\tan\theta > 0$, find $\tan\theta$. 2. **Rewrite the equation:** We have $$7\cos\theta - \sin\theta = 5.$$ 3. **Express in terms of sine and cosine:** Recall that $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\tan\theta > 0$ means sine and cosine have the same sign. 4. **Use the identity:** Let $R = \sqrt{7^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}$. 5. **Rewrite the left side as a single cosine:** $$7\cos\theta - \sin\theta = R \cos(\theta + \alpha),$$ where $\cos\alpha = \frac{7}{R} = \frac{7}{5\sqrt{2}}$ and $\sin\alpha = \frac{1}{R} = \frac{1}{5\sqrt{2}}$ (note the minus sign in front of $\sin\theta$ means $\alpha$ is positive). 6. **Solve for $\cos(\theta + \alpha)$:** $$R \cos(\theta + \alpha) = 5 \implies \cos(\theta + \alpha) = \frac{5}{R} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}.$$ 7. **Find $\theta + \alpha$:** $$\cos(\theta + \alpha) = \frac{1}{\sqrt{2}} \implies \theta + \alpha = \frac{\pi}{4} \text{ or } \theta + \alpha = \frac{7\pi}{4}.$$ 8. **Determine the correct angle using $\tan\theta > 0$:** Since $\tan\theta > 0$, $\theta$ is in the first or third quadrant. Calculate $\alpha$: $$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{1/(5\sqrt{2})}{7/(5\sqrt{2})} = \frac{1}{7}.$$ So, $\alpha = \arctan\left(\frac{1}{7}\right)$, which is a small positive angle. 9. **Check $\theta$ values:** - If $\theta + \alpha = \frac{\pi}{4}$, then $$\theta = \frac{\pi}{4} - \alpha,$$ which is positive and in the first quadrant. - If $\theta + \alpha = \frac{7\pi}{4}$, then $$\theta = \frac{7\pi}{4} - \alpha,$$ which is in the fourth quadrant, where $\tan\theta < 0$, so discard. 10. **Calculate $\tan\theta$:** $$\tan\theta = \tan\left(\frac{\pi}{4} - \alpha\right) = \frac{\tan\frac{\pi}{4} - \tan\alpha}{1 + \tan\frac{\pi}{4} \tan\alpha} = \frac{1 - \frac{1}{7}}{1 + 1 \cdot \frac{1}{7}} = \frac{\frac{6}{7}}{\frac{8}{7}} = \frac{6}{8} = \frac{3}{4}.$$ **Final answer:** $$\boxed{\tan\theta = \frac{3}{4}}.$$