If 2 tan A = 3 …

2x+y=7 4x-3y+1=0

Given that \(2 \tan A = 3\), we want to find the value of \(\sin A\) and \(\cos A\). 1. **Express \(\tan A\) in terms of a simpler fraction**: \[ \tan A = \frac{3}{2} \] 2. **Use the Pythagorean identity to find \(\sin A\) and \(\cos A\)**: We know that: \[ \sin^2 A + \cos^2 A = 1 \] and \[ \tan A = \frac{\sin A}{\cos A} \] Since \(\tan A = \frac{3}{2}\), we can set: \[ \sin A = 3k \quad \text{and} \quad \cos A = 2k \] where \(k\) is a common factor. 3. **Use the Pythagorean identity** to find \(k\): \[ \sin^2 A + \cos^2 A = 1 \] Substitute \(\sin A\) and \(\cos A\): \[ (3k)^2 + (2k)^2 = 1 \] Simplify the equation: \[ 9k^2 + 4k^2 = 1 \] \[ 13k^2 = 1 \] Solve for \(k\): \[ k^2 = \frac{1}{13} \] \[ k = \frac{1}{\sqrt{13}} \] 4. **Find \(\sin A\) and \(\cos A\)** using \(k\): \[ \sin A = 3k = 3 \times \frac{1}{\sqrt{13}} = \frac{3}{\sqrt{13}} = \frac{3 \sqrt{13}}{13} \] \[ \cos A = 2k = 2 \times \frac{1}{\sqrt{13}} = \frac{2}{\sqrt{13}} = \frac{2 \sqrt{13}}{13} \] Thus, the values of \(\sin A\) and \(\cos A\) are: \[ \sin A = \frac{3 \sqrt{13}}{13} \] \[ \cos A = \frac{2 \sqrt{13}}{13} \]