If `tantheta=sqrt(n),` where `n in N ,geq2,t h e nsec2theta` is always a rational number (b) an irrational number a positive integer (d) a negative integer

Options

A. a rational number
B. an irrational number
C. a positive integer
D. a negative integer.

Explanation

To solve the problem, we need to analyze the relationship between \( \tan \theta \) and \( \sec^2 \theta \).

Given:
\[ \tan \theta = \sqrt{n} \]
where \( n \in \mathbb{N} \) and \( n \geq 2 \).

### Step 1: Use the identity relating \( \tan \theta \) and \( \sec^2 \theta \)
We know the trigonometric identity:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]

### Step 2: Substitute \( \tan \theta \) into the identity
Substituting \( \tan \theta = \sqrt{n} \) into the identity gives:
\[ \sec^2 \theta = 1 + (\sqrt{n})^2 \]

### Step 3: Simplify the expression
Now, simplifying the equation:
\[ \sec^2 \theta = 1 + n \]

### Step 4: Analyze the result
Since \( n \) is a natural number and \( n \geq 2 \), we can conclude that:
\[ 1 + n \geq 1 + 2 = 3 \]
Thus, \( \sec^2 \theta \) is always greater than or equal to 3.

### Step 5: Determine the nature of \( \sec^2 \theta \)
Since \( n \) is a natural number, \( 1 + n \) is also a natural number. Therefore, \( \sec^2 \theta \) is a positive integer.

### Conclusion
The answer is that \( \sec^2 \theta \) is always a positive integer.