If `y=tan^(2)theta+sectheta, thetane(2n+1)pi//2,then`

Options

A. `y in(-oo,1]`
B. `yin(-oo,-1]`
C. `yin[-1,oo)`
D. none of these

Explanation

To find the range of the function \( y = \tan^2 \theta + \sec \theta \) given that \( \theta \neq \frac{(2n+1)\pi}{2} \), we can follow these steps:

### Step 1: Rewrite the function using trigonometric identities
We know that:
\[ 1 + \tan^2 \theta = \sec^2 \theta \]
Thus, we can express \( \tan^2 \theta \) as:
\[ \tan^2 \theta = \sec^2 \theta - 1 \]
Substituting this into the equation for \( y \):
\[ y = \tan^2 \theta + \sec \theta = (\sec^2 \theta - 1) + \sec \theta \]
This simplifies to:
\[ y = \sec^2 \theta + \sec \theta - 1 \]

### Step 2: Rearranging the equation
We can rearrange the equation as follows:
\[ y = \sec^2 \theta + \sec \theta - 1 \]
This can be treated as a quadratic equation in terms of \( \sec \theta \):
\[ \sec^2 \theta + \sec \theta - (y + 1) = 0 \]

### Step 3: Identifying the coefficients
In the quadratic equation \( ax^2 + bx + c = 0 \), we have:
- \( a = 1 \)
- \( b = 1 \)
- \( c = -(y + 1) \)

### Step 4: Applying the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ \sec \theta = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-(y + 1))}}{2 \cdot 1} \]
This simplifies to:
\[ \sec \theta = \frac{-1 \pm \sqrt{1 + 4y + 4}}{2} = \frac{-1 \pm \sqrt{4y + 5}}{2} \]

### Step 5: Finding the conditions for real values
For \( \sec \theta \) to be real, the expression under the square root must be non-negative:
\[ 4y + 5 \geq 0 \]
This leads to:
\[ y \geq -\frac{5}{4} \]

### Step 6: Analyzing the range of \( \sec \theta \)
We know that \( \sec \theta \) can take values \( \sec \theta \leq -1 \) or \( \sec \theta \geq 1 \). Thus, we need to analyze the two cases:

1. **Case 1**: \( \sec \theta \leq -1 \)
\[ \frac{-1 - \sqrt{4y + 5}}{2} \leq -1 \]
This leads to:
\[ -1 - \sqrt{4y + 5} \leq -2 \implies \sqrt{4y + 5} \geq 1 \implies 4y + 5 \geq 1 \implies 4y \geq -4 \implies y \geq -1 \]

2. **Case 2**: \( \sec \theta \geq 1 \)
\[ \frac{-1 + \sqrt{4y + 5}}{2} \geq 1 \]
This leads to:
\[ -1 + \sqrt{4y + 5} \geq 2 \implies \sqrt{4y + 5} \geq 3 \implies 4y + 5 \geq 9 \implies 4y \geq 4 \implies y \geq 1 \]

### Step 7: Conclusion
From both cases, we find that \( y \) must satisfy:
\[ y \geq -1 \quad \text{and} \quad y \geq 1 \]
Thus, the overall range of \( y \) is:
\[ y \in [-1, \infty) \]

### Final Answer
The range of \( y \) is \( [-1, \infty) \).

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