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

Options

A. `(1- e ^(2)) ^(3//2)`
B. `(2 -e ^(2)) ^(1//2)`
C. `(2- e ^(2)) ^(3//2)`
D. None of these

Answer

To solve the problem, we start with the given equation:

**Given:**
\[ \tan^2 \theta = 1 - e^2 \]

We need to find:
\[ \sec \theta + \tan^3 \theta \csc \theta \]

### Step 1: Find \(\sec^2 \theta\)

Using the identity:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
Substituting the value of \(\tan^2 \theta\):
\[ \sec^2 \theta = 1 + (1 - e^2) = 2 - e^2 \]

### Step 2: Find \(\sec \theta\)

Taking the square root of both sides, we find:
\[ \sec \theta = \sqrt{2 - e^2} \]

### Step 3: Find \(\tan \theta\)

Since \(\tan^2 \theta = 1 - e^2\), we can find \(\tan \theta\):
\[ \tan \theta = \sqrt{1 - e^2} \]

### Step 4: Find \(\csc \theta\)

Using the identity:
\[ \csc \theta = \frac{1}{\sin \theta} \]
We know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Thus:
\[ \sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 - e^2}{2 - e^2} \]
So:
\[ \sin \theta = \sqrt{\frac{1 - e^2}{2 - e^2}} \]
Thus:
\[ \csc \theta = \frac{1}{\sin \theta} = \sqrt{\frac{2 - e^2}{1 - e^2}} \]

### Step 5: Find \(\tan^3 \theta\)

Calculating \(\tan^3 \theta\):
\[ \tan^3 \theta = (\tan \theta)^3 = (1 - e^2)^{3/2} \]

### Step 6: Substitute into the expression

Now substituting back into the expression:
\[ \sec \theta + \tan^3 \theta \csc \theta = \sqrt{2 - e^2} + (1 - e^2)^{3/2} \cdot \sqrt{\frac{2 - e^2}{1 - e^2}} \]

### Step 7: Simplify the expression

This simplifies to:
\[ \sec \theta + \tan^3 \theta \csc \theta = \sqrt{2 - e^2} + \frac{(1 - e^2)^{3/2} \sqrt{2 - e^2}}{\sqrt{1 - e^2}} \]
\[ = \sqrt{2 - e^2} + \sqrt{2 - e^2} (1 - e^2) \]
\[ = \sqrt{2 - e^2} (1 + (1 - e^2)) = \sqrt{2 - e^2} (2 - e^2) \]

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