If sec `theta` is the eccentricity of a hyper bola then the eccentricity of the conjugate hyperbola is

Options

A. `tan theta`
B. `cot theta`
C. `cos theta`
D. `cosec theta`

Answer

To find the eccentricity of the conjugate hyperbola given that the eccentricity of the hyperbola is \( e_1 = \sec \theta \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between the eccentricities of hyperbolas:**
The relationship between the eccentricities of a hyperbola and its conjugate hyperbola is given by the formula:
\[ \frac{1}{e_1^2} + \frac{1}{e_2^2} = 1 \]
where \( e_1 \) is the eccentricity of the hyperbola and \( e_2 \) is the eccentricity of the conjugate hyperbola.

2. **Substitute the value of \( e_1 \):**
We know that \( e_1 = \sec \theta \). Therefore, we can substitute this into the equation:
\[ \frac{1}{(\sec \theta)^2} + \frac{1}{e_2^2} = 1 \]

3. **Simplify \( \frac{1}{(\sec \theta)^2} \):**
Recall that \( \sec \theta = \frac{1}{\cos \theta} \), so:
\[ \frac{1}{(\sec \theta)^2} = \cos^2 \theta \]
Thus, the equation becomes:
\[ \cos^2 \theta + \frac{1}{e_2^2} = 1 \]

4. **Rearrange the equation to solve for \( \frac{1}{e_2^2} \):**
Rearranging gives us:
\[ \frac{1}{e_2^2} = 1 - \cos^2 \theta \]

5. **Use the Pythagorean identity:**
We know from trigonometry that:
\[ 1 - \cos^2 \theta = \sin^2 \theta \]
Therefore, we can rewrite the equation as:
\[ \frac{1}{e_2^2} = \sin^2 \theta \]

6. **Reciprocate to find \( e_2^2 \):**
Taking the reciprocal gives:
\[ e_2^2 = \frac{1}{\sin^2 \theta} \]

7. **Take the square root to find \( e_2 \):**
Finally, taking the square root yields:
\[ e_2 = \csc \theta \]

### Conclusion:
Thus, the eccentricity of the conjugate hyperbola is:
\[ e_2 = \csc \theta \]