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# If 2 tan A = 3 …

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If 2 tan A = 3 , then the value of

Yuvraj Shinde 1 month, 2 weeks ago

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

I Am Helper 1 month, 2 weeks ago

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

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