If `tanalpha=sqrta` where a is not a perfect square then which of the following is a rational number

Options

A. `sin 2 alpha`
B. `tan 2 alpha`
C. `cos 2 alpha`
D. none of these

Explanation

To solve the problem, we need to determine which of the given expressions is a rational number when \( \tan \alpha = \sqrt{a} \) and \( a \) is not a perfect square.

### Step-by-Step Solution:

1. **Understand the given information**:
We have \( \tan \alpha = \sqrt{a} \), where \( a \) is not a perfect square. This means that \( \sqrt{a} \) is an irrational number.

2. **Evaluate the first option: \( \sin 2\alpha \)**:
The formula for \( \sin 2\alpha \) is:
\[ \sin 2\alpha = \frac{2 \tan \alpha}{1 + \tan^2 \alpha} \]
Substituting \( \tan \alpha = \sqrt{a} \):
\[ \sin 2\alpha = \frac{2 \sqrt{a}}{1 + a} \]
Since \( \sqrt{a} \) is irrational, \( \sin 2\alpha \) is also irrational. Thus, the first option is incorrect.

3. **Evaluate the second option: \( \tan 2\alpha \)**:
The formula for \( \tan 2\alpha \) is:
\[ \tan 2\alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} \]
Substituting \( \tan \alpha = \sqrt{a} \):
\[ \tan 2\alpha = \frac{2 \sqrt{a}}{1 - a} \]
Again, since \( \sqrt{a} \) is irrational, \( \tan 2\alpha \) is also irrational. Thus, the second option is incorrect.

4. **Evaluate the third option: \( \cos 2\alpha \)**:
The formula for \( \cos 2\alpha \) is:
\[ \cos 2\alpha = \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} \]
Substituting \( \tan \alpha = \sqrt{a} \):
\[ \cos 2\alpha = \frac{1 - a}{1 + a} \]
Here, both \( 1 - a \) and \( 1 + a \) are integers, and the division of two integers (where the denominator is not zero) results in a rational number. Thus, the third option is correct.

### Conclusion:
The rational number among the options is \( \cos 2\alpha = \frac{1 - a}{1 + a} \).