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: $$\sin A = \frac{3 \sqrt{13}}{13}$$ $$\cos A = \frac{2 \sqrt{13}}{13}$$