If `tan^2 theta =(1-e^2)`, then sec `theta+tan^3 theta "cosec" theta` is equal to

Options

A. `(2+e^2)^(3//2)`
B. `(2-e^2)^(3//2)`
C. `(1-e^2)^(3//2)`
D. `(1+e^2)^(3//2)`

Answer

To solve the problem, we need to find the value of \( \sec \theta + \tan^3 \theta \csc \theta \) given that \( \tan^2 \theta = 1 - e^2 \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \tan^2 \theta = 1 - e^2 \]

2. **Use the identity for secant:**
We know that:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
Substituting the value of \( \tan^2 \theta \):
\[ \sec^2 \theta = 1 + (1 - e^2) = 2 - e^2 \]

3. **Find \( \sec \theta \):**
Taking the square root:
\[ \sec \theta = \sqrt{2 - e^2} \]

4. **Calculate \( \tan \theta \):**
Since \( \tan^2 \theta = 1 - e^2 \), we take the square root:
\[ \tan \theta = \sqrt{1 - e^2} \]

5. **Calculate \( \tan^3 \theta \):**
\[ \tan^3 \theta = (\tan \theta)^3 = (\sqrt{1 - e^2})^3 = (1 - e^2)^{3/2} \]

6. **Find \( \csc \theta \):**
We know that:
\[ \csc \theta = \frac{1}{\sin \theta} \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) and knowing \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ \sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 - e^2}{2 - e^2} \]
Thus,
\[ \csc \theta = \frac{1}{\sqrt{\frac{1 - e^2}{2 - e^2}}} = \sqrt{\frac{2 - e^2}{1 - e^2}} \]

7. **Combine the terms:**
Now we can substitute back into the original expression:
\[ \sec \theta + \tan^3 \theta \csc \theta = \sqrt{2 - e^2} + (1 - e^2)^{3/2} \cdot \sqrt{\frac{2 - e^2}{1 - e^2}} \]

8. **Simplify:**
The expression can be simplified further, but we can also express it as:
\[ \sec \theta + \tan^3 \theta \csc \theta = \sqrt{2 - e^2} + \frac{(1 - e^2)^{3/2} \sqrt{2 - e^2}}{\sqrt{1 - e^2}} \]
This leads to:
\[ = \sqrt{2 - e^2} \left( 1 + \frac{(1 - e^2)}{\sqrt{1 - e^2}} \right) = \sqrt{2 - e^2} \left( 1 + \sqrt{1 - e^2} \right) \]

9. **Final Result:**
The final expression can be simplified to:
\[ (2 - e^2)^{3/2} \]

### Final Answer:
\[ (2 - e^2)^{3/2} \]