If ` cos ^(2) theta = 3 sin ^(2) theta , ` then find the value of ` sec ^(2) theta - tan ^(2) theta + cos ^(2) theta`

Options

A. ` 7//4`
B. ` 5//2`
C. ` 8//9`
D. ` 3//5`

Answer

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

1. **Given**: \( \cos^2 \theta = 3 \sin^2 \theta \)

2. **Using the Pythagorean Identity**: We know that \( \sin^2 \theta + \cos^2 \theta = 1 \). We can express \( \sin^2 \theta \) in terms of \( \cos^2 \theta \):
\[ \sin^2 \theta = 1 - \cos^2 \theta \]

3. **Substituting in the given equation**:
\[ \cos^2 \theta = 3(1 - \cos^2 \theta) \]
Expanding this gives:
\[ \cos^2 \theta = 3 - 3\cos^2 \theta \]

4. **Rearranging the equation**:
\[ \cos^2 \theta + 3\cos^2 \theta = 3 \]
This simplifies to:
\[ 4\cos^2 \theta = 3 \]

5. **Solving for \( \cos^2 \theta \)**:
\[ \cos^2 \theta = \frac{3}{4} \]

6. **Finding \( \sec^2 \theta \) and \( \tan^2 \theta \)**:
- Recall that \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \):
\[ \sec^2 \theta = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]
- And \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \):
\[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{3}{4} = \frac{1}{4} \]
Thus,
\[ \tan^2 \theta = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \]

7. **Calculating \( \sec^2 \theta - \tan^2 \theta + \cos^2 \theta \)**:
\[ \sec^2 \theta - \tan^2 \theta + \cos^2 \theta = \frac{4}{3} - \frac{1}{3} + \frac{3}{4} \]

8. **Finding a common denominator**: The common denominator for \( 3 \) and \( 4 \) is \( 12 \):
\[ \frac{4}{3} = \frac{16}{12}, \quad \frac{1}{3} = \frac{4}{12}, \quad \frac{3}{4} = \frac{9}{12} \]

9. **Combining the fractions**:
\[ \sec^2 \theta - \tan^2 \theta + \cos^2 \theta = \frac{16}{12} - \frac{4}{12} + \frac{9}{12} = \frac{16 - 4 + 9}{12} = \frac{21}{12} = \frac{7}{4} \]

Thus, the final answer is:
\[ \boxed{\frac{7}{4}} \]