If ` cos theta= (2sqrt(mn))/(m+n) ` , find the value of ` sin theta ` (given m > n )

Answer

To find the value of \( \sin \theta \) given that \( \cos \theta = \frac{2\sqrt{mn}}{m+n} \) and \( m > n \), we can use the Pythagorean identity:

### Step-by-Step Solution:

1. **Use the Pythagorean Identity**:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
From this, we can express \( \sin^2 \theta \) as:
\[ \sin^2 \theta = 1 - \cos^2 \theta \]

2. **Substitute the value of \( \cos \theta \)**:
Substitute \( \cos \theta = \frac{2\sqrt{mn}}{m+n} \) into the equation:
\[ \sin^2 \theta = 1 - \left(\frac{2\sqrt{mn}}{m+n}\right)^2 \]

3. **Calculate \( \cos^2 \theta \)**:
\[ \cos^2 \theta = \left(\frac{2\sqrt{mn}}{m+n}\right)^2 = \frac{4mn}{(m+n)^2} \]
Now substitute this back into the equation for \( \sin^2 \theta \):
\[ \sin^2 \theta = 1 - \frac{4mn}{(m+n)^2} \]

4. **Find a common denominator**:
Rewrite 1 as \( \frac{(m+n)^2}{(m+n)^2} \):
\[ \sin^2 \theta = \frac{(m+n)^2 - 4mn}{(m+n)^2} \]

5. **Simplify the numerator**:
Expand \( (m+n)^2 \):
\[ (m+n)^2 = m^2 + 2mn + n^2 \]
Therefore,
\[ \sin^2 \theta = \frac{m^2 + 2mn + n^2 - 4mn}{(m+n)^2} = \frac{m^2 - 2mn + n^2}{(m+n)^2} \]

6. **Recognize the numerator as a square**:
The numerator can be factored as:
\[ m^2 - 2mn + n^2 = (m-n)^2 \]
Thus,
\[ \sin^2 \theta = \frac{(m-n)^2}{(m+n)^2} \]

7. **Take the square root**:
To find \( \sin \theta \), take the square root of both sides:
\[ \sin \theta = \frac{m-n}{m+n} \]

8. **Consider the sign**:
Since \( m > n \), \( m - n \) is positive. Therefore, we take the positive root:
\[ \sin \theta = \frac{m-n}{m+n} \]

### Final Answer:
\[ \sin \theta = \frac{m-n}{m+n} \]