If `tantheta = (3)/( 4)` find the value of `((1-cos^(2)theta)/(1+cos^(2)theta))`

Answer

To solve the problem, we need to find the value of the expression \(\frac{1 - \cos^2 \theta}{1 + \cos^2 \theta}\) given that \(\tan \theta = \frac{3}{4}\).

### Step-by-Step Solution:

1. **Use the identity for sine and cosine**:
We know that \(\sin^2 \theta + \cos^2 \theta = 1\). From this, we can express \(\sin^2 \theta\) as:
\[ \sin^2 \theta = 1 - \cos^2 \theta \]
Therefore, we can rewrite the expression:
\[ \frac{1 - \cos^2 \theta}{1 + \cos^2 \theta} = \frac{\sin^2 \theta}{1 + \cos^2 \theta} \]

2. **Find \(\sin \theta\) and \(\cos \theta\) using \(\tan \theta\)**:
Given \(\tan \theta = \frac{3}{4}\), we can consider a right triangle where the opposite side is 3 and the adjacent side is 4. The hypotenuse \(h\) can be calculated using the Pythagorean theorem:
\[ h = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
Now, we can find \(\sin \theta\) and \(\cos \theta\):
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}, \quad \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5} \]

3. **Calculate \(\sin^2 \theta\) and \(\cos^2 \theta\)**:
\[ \sin^2 \theta = \left(\frac{3}{5}\right)^2 = \frac{9}{25}, \quad \cos^2 \theta = \left(\frac{4}{5}\right)^2 = \frac{16}{25} \]

4. **Substitute \(\sin^2 \theta\) and \(\cos^2 \theta\) into the expression**:
Now substitute \(\sin^2 \theta\) and \(\cos^2 \theta\) into the expression we derived:
\[ \frac{\sin^2 \theta}{1 + \cos^2 \theta} = \frac{\frac{9}{25}}{1 + \frac{16}{25}} = \frac{\frac{9}{25}}{\frac{25 + 16}{25}} = \frac{\frac{9}{25}}{\frac{41}{25}} \]

5. **Simplify the expression**:
Dividing by a fraction is the same as multiplying by its reciprocal:
\[ = \frac{9}{25} \times \frac{25}{41} = \frac{9}{41} \]

### Final Answer:
Thus, the value of \(\frac{1 - \cos^2 \theta}{1 + \cos^2 \theta}\) is \(\frac{9}{41}\).