If `sec ^(2) theta + tan ^(2) theta = 5/3,` what is the value of tan `2 theta` ?

Options

A. `2sqrt3`
B. `sqrt3`
C. `(1)/(sqrt3)`
D. cannot be detrmined

Answer

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

1. **Given Equation**:
\[ \sec^2 \theta + \tan^2 \theta = \frac{5}{3} \]

2. **Using the Identity**:
We know the trigonometric identity:
\[ \sec^2 \theta - \tan^2 \theta = 1 \]
Let's denote:
\[ x = \sec^2 \theta \quad \text{and} \quad y = \tan^2 \theta \]
Thus, we can rewrite our equations as:
\[ x + y = \frac{5}{3} \quad \text{(1)} \]
\[ x - y = 1 \quad \text{(2)} \]

3. **Adding the Two Equations**:
Now, we can add equations (1) and (2):
\[ (x + y) + (x - y) = \frac{5}{3} + 1 \]
Simplifying this gives:
\[ 2x = \frac{5}{3} + \frac{3}{3} = \frac{8}{3} \]
Therefore:
\[ x = \frac{4}{3} \]

4. **Substituting Back to Find y**:
Now, substitute \( x \) back into equation (1):
\[ \frac{4}{3} + y = \frac{5}{3} \]
Solving for \( y \):
\[ y = \frac{5}{3} - \frac{4}{3} = \frac{1}{3} \]

5. **Finding tan 2θ**:
We know that:
\[ \tan^2 \theta = y = \frac{1}{3} \]
Now, we can use the double angle formula for tangent:
\[ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \]
First, we find \( \tan \theta \):
\[ \tan \theta = \sqrt{\tan^2 \theta} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \]
Now substituting into the double angle formula:
\[ \tan 2\theta = \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{2/\sqrt{3}}{2/3} = \frac{2/\sqrt{3} \cdot 3/2} = \frac{3}{\sqrt{3}} = \sqrt{3} \]

6. **Final Answer**:
Therefore, the value of \( \tan 2\theta \) is:
\[ \tan 2\theta = \sqrt{3} \]