If `sec2 theta=p+tan2 theta,` then the value of `sin^(2)theta` in therms of p is given by

Options

A. `((p-1)^(2))/(2(p^(2)+1))`
B. `1/2((p-1)/(p+1))^(2)`
C. `(p^(2)-1)/(1(p^(2)+1))`
D. `(p^(2)-1)/(2(p+1)^(2))`

Explanation

To find the value of \( \sin^2 \theta \) in terms of \( p \) given the equation \( \sec 2\theta = p + \tan 2\theta \), we can follow these steps:

### Step-by-step Solution:

1. **Start with the given equation:**
\[ \sec 2\theta = p + \tan 2\theta \]

2. **Rearrange the equation:**
\[ \sec 2\theta - \tan 2\theta = p \]

3. **Use the identity:**
\[ \sec^2 2\theta - \tan^2 2\theta = 1 \]
This can be factored as:
\[ (\sec 2\theta - \tan 2\theta)(\sec 2\theta + \tan 2\theta) = 1 \]

4. **Substitute the rearranged equation into the identity:**
Let \( x = \sec 2\theta - \tan 2\theta \). Then:
\[ x \cdot (\sec 2\theta + \tan 2\theta) = 1 \]
Since \( x = p \), we have:
\[ p \cdot (\sec 2\theta + \tan 2\theta) = 1 \]

5. **Solve for \( \sec 2\theta + \tan 2\theta \):**
\[ \sec 2\theta + \tan 2\theta = \frac{1}{p} \]

6. **Now we have two equations:**
\[ \sec 2\theta - \tan 2\theta = p \quad \text{(1)} \]
\[ \sec 2\theta + \tan 2\theta = \frac{1}{p} \quad \text{(2)} \]

7. **Add equations (1) and (2):**
\[ 2\sec 2\theta = p + \frac{1}{p} \]
\[ \sec 2\theta = \frac{p + \frac{1}{p}}{2} \]

8. **Recall that \( \sec 2\theta = \frac{1}{\cos 2\theta} \):**
\[ \cos 2\theta = \frac{2}{p + \frac{1}{p}} = \frac{2p}{p^2 + 1} \]

9. **Use the double angle identity for cosine:**
\[ \cos 2\theta = 1 - 2\sin^2 \theta \]
Setting the two expressions for \( \cos 2\theta \) equal gives:
\[ 1 - 2\sin^2 \theta = \frac{2p}{p^2 + 1} \]

10. **Rearranging gives:**
\[ 2\sin^2 \theta = 1 - \frac{2p}{p^2 + 1} \]

11. **Combine the fractions:**
\[ 2\sin^2 \theta = \frac{(p^2 + 1) - 2p}{p^2 + 1} = \frac{p^2 - 2p + 1}{p^2 + 1} \]

12. **Recognize the numerator as a perfect square:**
\[ 2\sin^2 \theta = \frac{(p - 1)^2}{p^2 + 1} \]

13. **Finally, solve for \( \sin^2 \theta \):**
\[ \sin^2 \theta = \frac{(p - 1)^2}{2(p^2 + 1)} \]

### Final Answer:
\[ \sin^2 \theta = \frac{(p - 1)^2}{2(p^2 + 1)} \]