If `sec theta- tan theta= P`, then `cosec theta`= ?
यदि `sec theta- tan theta= P`, है, तो `cosec theta`= ?

Options

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

Answer

To solve the problem, we need to find the value of \( \csc \theta \) given that \( \sec \theta - \tan \theta = P \).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ \sec \theta - \tan \theta = P \]

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

3. **Substitute the value of \( P \)**:
Substitute \( \sec \theta - \tan \theta \) with \( P \):
\[ P(\sec \theta + \tan \theta) = 1 \]

4. **Solve for \( \sec \theta + \tan \theta \)**:
Rearranging gives:
\[ \sec \theta + \tan \theta = \frac{1}{P} \]

5. **Now we have two equations**:
- \( \sec \theta - \tan \theta = P \) (Equation 1)
- \( \sec \theta + \tan \theta = \frac{1}{P} \) (Equation 2)

6. **Add the two equations**:
Adding Equation 1 and Equation 2:
\[ (\sec \theta - \tan \theta) + (\sec \theta + \tan \theta) = P + \frac{1}{P} \]
This simplifies to:
\[ 2 \sec \theta = P + \frac{1}{P} \]

7. **Solve for \( \sec \theta \)**:
\[ \sec \theta = \frac{P + \frac{1}{P}}{2} \]

8. **Find \( \cos \theta \)**:
Since \( \sec \theta = \frac{1}{\cos \theta} \), we have:
\[ \cos \theta = \frac{2}{P + \frac{1}{P}} \]

9. **Use the identity for \( \sin \theta \)**:
We know:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
So,
\[ \sin^2 \theta = 1 - \cos^2 \theta \]

10. **Substitute \( \cos \theta \)**:
\[ \sin^2 \theta = 1 - \left(\frac{2}{P + \frac{1}{P}}\right)^2 \]

11. **Calculate \( \csc \theta \)**:
Since \( \csc \theta = \frac{1}{\sin \theta} \), we need to find \( \sin \theta \) first:
\[ \sin \theta = \sqrt{1 - \left(\frac{2}{P + \frac{1}{P}}\right)^2} \]
Thus,
\[ \csc \theta = \frac{1}{\sqrt{1 - \left(\frac{2}{P + \frac{1}{P}}\right)^2}} \]

### Final Answer:
\[ \csc \theta = \frac{1}{\sqrt{1 - \left(\frac{2}{P + \frac{1}{P}}\right)^2}} \]