If \( 7 \tan \theta=3 \) and \( \cos \theta>0 \), find the value of \( \frac{\sin \theta+\cos \theta}{2 \sin \theta} \)

Given the equation \( 7 \tan \theta = 3 \), we can express \(\tan \theta\) in terms of sine and cosine: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] Thus, we have: \[ 7 \frac{\sin \theta}{\cos \theta} = 3 \implies \sin \theta = \frac{3}{7} \cos \theta \] Next, we will use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\). Substitute \(\sin \theta\) from the equation above: \[ \left(\frac{3}{7} \cos \theta\right)^2 + \cos^2 \theta = 1 \] This simplifies to: \[ \frac{9}{49} \cos^2 \theta + \cos^2 \theta = 1 \] Combining like terms gives: \[ \left(\frac{9}{49} + 1\right) \cos^2 \theta = 1 \] \[ \left(\frac{9}{49} + \frac{49}{49}\right) \cos^2 \theta = 1 \implies \frac{58}{49} \cos^2 \theta = 1 \] To solve for \(\cos^2 \theta\), we multiply by \(\frac{49}{58}\): \[ \cos^2 \theta = \frac{49}{58} \] Now, take the square root to find \(\cos \theta\) (we know \(\cos \theta > 0\)): \[ \cos \theta = \sqrt{\frac{49}{58}} = \frac{7}{\sqrt{58}} \] Next, we will calculate \(\sin \theta\) using \(\sin \theta = \frac{3}{7} \cos \theta\): \[ \sin \theta = \frac{3}{7} \cdot \frac{7}{\sqrt{58}} = \frac{3}{\sqrt{58}} \] Now, we have: \[ \sin \theta = \frac{3}{\sqrt{58}}, \quad \cos \theta = \frac{7}{\sqrt{58}} \] We want to find the value of: \[ \frac{\sin \theta + \cos \theta}{2 \sin \theta} \] Substituting \(\sin \theta\) and \(\cos \theta\): \[ \frac{\frac{3}{\sqrt{58}} + \frac{7}{\sqrt{58}}}{2 \cdot \frac{3}{\sqrt{58}}} = \frac{\frac{10}{\sqrt{58}}}{\frac{6}{\sqrt{58}}} = \frac{10}{6} = \frac{5}{3} \] Thus, the value is: \[ \boxed{\frac{5}{3}} \]