If `tanalpha=sqrt(a)`, where a is a rational number whch is not a perfect square, then which of the following is a rational number?

Options

A. `sin2alpha`
B. `tan2alpha`
C. `cos2alpha`
D. `" cosec "2alpha`

Answer

To solve the problem, we need to analyze the given information and apply trigonometric identities to determine which of the provided options is a rational number.

### Given:
- \( \tan \alpha = \sqrt{a} \) where \( a \) is a rational number that is not a perfect square.

### We need to find:
- Which of the following expressions is a rational number:
1. \( \sin 2\alpha \)
2. \( \tan 2\alpha \)
3. \( \cos 2\alpha \)
4. \( \csc 2\alpha \)

### Step-by-step Solution:

1. **Calculate \( \sin 2\alpha \)**:
\[ \sin 2\alpha = \frac{2 \tan \alpha}{1 + \tan^2 \alpha} \]
Substituting \( \tan \alpha = \sqrt{a} \):
\[ \sin 2\alpha = \frac{2 \sqrt{a}}{1 + a} \]
Since \( \sqrt{a} \) is not a rational number (as \( a \) is not a perfect square), \( \sin 2\alpha \) is not a rational number.

2. **Calculate \( \tan 2\alpha \)**:
\[ \tan 2\alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} \]
Substituting \( \tan \alpha = \sqrt{a} \):
\[ \tan 2\alpha = \frac{2 \sqrt{a}}{1 - a} \]
Again, since \( \sqrt{a} \) is not a rational number, \( \tan 2\alpha \) is not a rational number.

3. **Calculate \( \cos 2\alpha \)**:
\[ \cos 2\alpha = \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} \]
Substituting \( \tan \alpha = \sqrt{a} \):
\[ \cos 2\alpha = \frac{1 - a}{1 + a} \]
Here, both \( 1 - a \) and \( 1 + a \) are rational numbers, and since the denominator is not zero (as \( a \) is a rational number), \( \cos 2\alpha \) is a rational number.

4. **Calculate \( \csc 2\alpha \)**:
\[ \csc 2\alpha = \frac{1}{\sin 2\alpha} \]
Since \( \sin 2\alpha \) is not a rational number, \( \csc 2\alpha \) is also not a rational number.

### Conclusion:
The only expression that is a rational number is \( \cos 2\alpha \).

### Final Answer:
**The rational number is \( \cos 2\alpha \) (Option 3).**

---