If the equation `"sec" theta + "cosec" theta =c` has real roots between 0 and `2pi`, then

Options

A. `c^(2) lt 8`
B. `c^(2) gt= 8`
C. `c^(2) = 8`
D. none of these

Answer

To solve the equation \( \sec \theta + \csc \theta = c \) for real roots between \( 0 \) and \( 2\pi \), we will follow these steps:

### Step 1: Rewrite the equation in terms of sine and cosine
We start by rewriting the secant and cosecant functions in terms of sine and cosine:
\[ \sec \theta = \frac{1}{\cos \theta}, \quad \csc \theta = \frac{1}{\sin \theta} \]
Thus, the equation becomes:
\[ \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = c \]

### Step 2: Combine the fractions
To combine the fractions, we find a common denominator:
\[ \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} = c \]
This can be rearranged to:
\[ \sin \theta + \cos \theta = c \sin \theta \cos \theta \]

### Step 3: Square both sides
Next, we square both sides to eliminate the sine and cosine terms:
\[ (\sin \theta + \cos \theta)^2 = (c \sin \theta \cos \theta)^2 \]
Expanding both sides gives:
\[ \sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta = c^2 \sin^2 \theta \cos^2 \theta \]
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can simplify this to:
\[ 1 + 2 \sin \theta \cos \theta = c^2 \sin^2 \theta \cos^2 \theta \]

### Step 4: Substitute \( \sin 2\theta \)
Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Thus, we can rewrite the equation as:
\[ 1 + \sin 2\theta = \frac{c^2}{4} \sin^2 2\theta \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ c^2 \sin^2 2\theta - \sin 2\theta - 4 = 0 \]
This is a quadratic equation in terms of \( \sin 2\theta \).

### Step 6: Apply the quadratic formula
Using the quadratic formula \( \sin 2\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = c^2 \), \( b = -1 \), and \( c = -4 \):
\[ \sin 2\theta = \frac{1 \pm \sqrt{1 + 16c^2}}{2c^2} \]

### Step 7: Determine the range of \( \sin 2\theta \)
Since \( \sin 2\theta \) must be between -1 and 1, we set up the inequalities:
\[ -1 \leq \frac{1 \pm \sqrt{1 + 16c^2}}{2c^2} \leq 1 \]

### Step 8: Solve the inequalities
1. For the upper bound:
\[ 1 + \sqrt{1 + 16c^2} \leq 2c^2 \implies \sqrt{1 + 16c^2} \leq 2c^2 - 1 \]
Squaring both sides gives:
\[ 1 + 16c^2 \leq 4c^4 - 4c^2 + 1 \implies 4c^4 - 20c^2 \geq 0 \implies 4c^2(c^2 - 5) \geq 0 \]
This implies \( c^2 \geq 5 \) or \( c^2 = 0 \).

2. For the lower bound:
\[ 1 - \sqrt{1 + 16c^2} \geq -2c^2 \implies \sqrt{1 + 16c^2} \leq 2c^2 + 1 \]
Squaring both sides gives:
\[ 1 + 16c^2 \leq 4c^4 + 4c^2 + 1 \implies 4c^4 - 12c^2 \geq 0 \implies 4c^2(c^2 - 3) \geq 0 \]
This implies \( c^2 \geq 3 \).

### Conclusion
Combining both inequalities, we find that \( c^2 \geq 5 \) is the necessary condition for the equation to have real roots between \( 0 \) and \( 2\pi \). Thus, the final result is:
\[ c \geq \sqrt{5} \]