If it is not possible to draw any tangent from the point (1/4, 1) to the parabola `y^(2)=4xcostheta+sin^(2)theta`, then `theta` belongs to

Options

A. `[-pi/2,pi/2]`
B. `[-pi/2,pi/2]-{0}`
C. `(-pi/2,pi/2)-{0}`
D. none of these

Answer

To solve the problem, we need to determine the values of \(\theta\) for which it is not possible to draw any tangent from the point \((\frac{1}{4}, 1)\) to the parabola given by the equation \(y^2 = 4x \cos \theta + \sin^2 \theta\).

### Step-by-Step Solution:

1. **Identify the Condition for Tangents**:
If a point lies inside the parabola, it means that no tangents can be drawn from that point to the parabola. Therefore, we need to check if the point \((\frac{1}{4}, 1)\) lies inside the parabola.

2. **Substitute the Point into the Parabola Equation**:
The equation of the parabola is \(y^2 = 4x \cos \theta + \sin^2 \theta\).
Substituting \(x = \frac{1}{4}\) and \(y = 1\):
\[ 1^2 = 4 \left(\frac{1}{4}\right) \cos \theta + \sin^2 \theta \]
This simplifies to:
\[ 1 = \cos \theta + \sin^2 \theta \]

3. **Rearranging the Equation**:
Rearranging the equation gives:
\[ 1 - \cos \theta - \sin^2 \theta = 0 \]
This can be rewritten as:
\[ \cos \theta + \sin^2 \theta = 1 \]

4. **Using the Pythagorean Identity**:
We know that \(\sin^2 \theta = 1 - \cos^2 \theta\). Substituting this into the equation gives:
\[ \cos \theta + (1 - \cos^2 \theta) = 1 \]
Simplifying this results in:
\[ \cos \theta - \cos^2 \theta = 0 \]

5. **Factoring the Equation**:
Factoring out \(\cos \theta\):
\[ \cos \theta (1 - \cos \theta) = 0 \]
This gives us two cases:
- \(\cos \theta = 0\)
- \(1 - \cos \theta = 0 \Rightarrow \cos \theta = 1\)

6. **Finding Values of \(\theta\)**:
- \(\cos \theta = 0\) occurs at \(\theta = \frac{\pi}{2} + n\pi\) for \(n \in \mathbb{Z}\).
- \(\cos \theta = 1\) occurs at \(\theta = 0 + 2n\pi\) for \(n \in \mathbb{Z}\).

7. **Determining the Range of \(\theta\)**:
Since we are looking for values of \(\theta\) such that the point \((\frac{1}{4}, 1)\) lies inside the parabola, we need to find the intervals where \(\cos \theta + \sin^2 \theta < 1\).

Analyzing the behavior of \(\sin^2 \theta\) (which is always non-negative and less than or equal to 1) and \(\cos \theta\), we find that:
- For \(\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})\), \(\cos \theta\) is positive and \(\sin^2 \theta\) is positive, thus the sum can be less than 1.
- At \(\theta = 0\), \(\cos \theta + \sin^2 \theta = 1\), which is not acceptable.

8. **Conclusion**:
Therefore, the values of \(\theta\) for which no tangent can be drawn from the point \((\frac{1}{4}, 1)\) to the parabola are:
\[ \theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \setminus \{0\} \]

### Final Answer:
\[ \theta \text{ belongs to } \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \text{ except } 0. \]