The eccentricity of hyperbola `x^2-y^2cosec^theta=5` is `sqrt7` times of eccentricity of ellipse `x^2+y^2cosec^theta`=5 then `theta` is where `0lt0ltpi/2`

Options

A. `theta=pi/3`
B. `theta=pi/2`
C. `theta=pi/(-3)`
D. `theta=(2pi)/3`

Answer

To solve the problem, we need to find the angle \( \theta \) such that the eccentricity of the hyperbola \( x^2 - y^2 \csc^2 \theta = 5 \) is \( \sqrt{7} \) times the eccentricity of the ellipse \( x^2 + y^2 \csc^2 \theta = 5 \).

### Step 1: Rewrite the equations in standard form

1. **Hyperbola**: The equation \( x^2 - y^2 \csc^2 \theta = 5 \) can be rewritten as:
\[ \frac{x^2}{5} - \frac{y^2}{5 \sin^2 \theta} = 1 \]
Here, \( a^2 = 5 \) and \( b^2 = 5 \sin^2 \theta \).

2. **Ellipse**: The equation \( x^2 + y^2 \csc^2 \theta = 5 \) can be rewritten as:
\[ \frac{x^2}{5} + \frac{y^2}{5 \sin^2 \theta} = 1 \]
Here, \( a^2 = 5 \) and \( b^2 = 5 \sin^2 \theta \).

### Step 2: Calculate the eccentricities

1. **Eccentricity of Hyperbola**:
The eccentricity \( E_1 \) of the hyperbola is given by:
\[ E_1 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{5 \sin^2 \theta}{5}} = \sqrt{1 + \sin^2 \theta} \]

2. **Eccentricity of Ellipse**:
The eccentricity \( E_2 \) of the ellipse is given by:
\[ E_2 = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{5 \sin^2 \theta}{5}} = \sqrt{1 - \sin^2 \theta} = \cos \theta \]

### Step 3: Set up the relationship between the eccentricities

According to the problem, the eccentricity of the hyperbola is \( \sqrt{7} \) times the eccentricity of the ellipse:
\[ E_1 = \sqrt{7} E_2 \]
Substituting the expressions for \( E_1 \) and \( E_2 \):
\[ \sqrt{1 + \sin^2 \theta} = \sqrt{7} \cos \theta \]

### Step 4: Square both sides

Squaring both sides gives:
\[ 1 + \sin^2 \theta = 7 \cos^2 \theta \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can replace \( \cos^2 \theta \):
\[ 1 + \sin^2 \theta = 7(1 - \sin^2 \theta) \]
Expanding and rearranging:
\[ 1 + \sin^2 \theta = 7 - 7 \sin^2 \theta \]
\[ 1 + 8 \sin^2 \theta = 7 \]
\[ 8 \sin^2 \theta = 6 \]
\[ \sin^2 \theta = \frac{3}{4} \]

### Step 5: Solve for \( \theta \)

Taking the square root:
\[ \sin \theta = \frac{\sqrt{3}}{2} \]
Since \( 0 < \theta < \frac{\pi}{2} \), we have:
\[ \theta = \frac{\pi}{3} \]

### Final Answer
Thus, the value of \( \theta \) is:
\[ \theta = \frac{\pi}{3} \]