If the pair of lines `b^2x^2-a^2y^2=0` are inclined at an angle `theta`, then the eccentricity of the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` is

Options

A. `sec theta`
B. `cos theta`
C. `tan""(theta)/(2)`
D. `sec""(theta)/(2)`

Answer

To solve the problem, we need to find the eccentricity of the hyperbola given that the pair of lines \( b^2x^2 - a^2y^2 = 0 \) are inclined at an angle \( \theta \).

### Step-by-Step Solution:

1. **Equation of the Pair of Lines**:
The equation \( b^2x^2 - a^2y^2 = 0 \) can be rewritten as:
\[ b^2x^2 = a^2y^2 \]
Taking the square root of both sides gives:
\[ y = \pm \frac{b}{a} x \]
This indicates that the pair of lines can be expressed as:
\[ y = \frac{b}{a} x \quad \text{and} \quad y = -\frac{b}{a} x \]

2. **Finding the Slopes**:
The slopes of the two lines are:
\[ m_1 = \frac{b}{a} \quad \text{and} \quad m_2 = -\frac{b}{a} \]

3. **Using the Angle Between Two Lines**:
The angle \( \theta \) between the two lines can be found using the formula:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]
Substituting \( m_1 \) and \( m_2 \):
\[ \tan \theta = \frac{\frac{b}{a} - \left(-\frac{b}{a}\right)}{1 + \left(\frac{b}{a}\right)\left(-\frac{b}{a}\right)} \]
Simplifying this gives:
\[ \tan \theta = \frac{\frac{2b}{a}}{1 - \frac{b^2}{a^2}} = \frac{2b/a}{(a^2 - b^2)/a^2} = \frac{2b a^2}{a^2 - b^2} \]

4. **Relating \( b/a \) to \( \theta \)**:
From the above, we can express:
\[ \tan \frac{\theta}{2} = \frac{b}{a} \]

5. **Finding the Eccentricity of the Hyperbola**:
The eccentricity \( e \) of the hyperbola given by the equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is calculated using the formula:
\[ e^2 = 1 + \frac{b^2}{a^2} \]
Substituting \( \frac{b}{a} = \tan \frac{\theta}{2} \):
\[ e^2 = 1 + \tan^2 \frac{\theta}{2} \]

6. **Using the Identity**:
We know from trigonometric identities that:
\[ 1 + \tan^2 x = \sec^2 x \]
Therefore:
\[ e^2 = \sec^2 \frac{\theta}{2} \]

7. **Final Result**:
Taking the square root gives:
\[ e = \sec \frac{\theta}{2} \]

### Conclusion:
The eccentricity of the hyperbola is:
\[ \boxed{\sec \frac{\theta}{2}} \]