The equation `tan^(4)x-2sec^(2)x+a^(2)=0` will have at least one solution, if

Options

A. `|a|le 4`
B. `|a|le 2`
C. `|a| le sqrt3`
D. `|a|gt 2`

Answer

To solve the equation \( \tan^4 x - 2 \sec^2 x + a^2 = 0 \) and determine the conditions under which it has at least one solution, we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \tan^4 x - 2 \sec^2 x + a^2 = 0 \]
Recall that \( \sec^2 x = 1 + \tan^2 x \). Therefore, we can express \( \sec^2 x \) in terms of \( \tan^2 x \):
\[ \tan^4 x - 2(1 + \tan^2 x) + a^2 = 0 \]

### Step 2: Simplify the equation
Expanding this gives:
\[ \tan^4 x - 2 - 2 \tan^2 x + a^2 = 0 \]
Rearranging terms, we have:
\[ \tan^4 x - 2 \tan^2 x + (a^2 - 2) = 0 \]

### Step 3: Let \( y = \tan^2 x \)
Let \( y = \tan^2 x \). The equation now becomes:
\[ y^2 - 2y + (a^2 - 2) = 0 \]

### Step 4: Apply the quadratic formula
To find the values of \( y \), we can use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2 \), and \( c = a^2 - 2 \):
\[ y = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (a^2 - 2)}}{2 \cdot 1} \]
This simplifies to:
\[ y = \frac{2 \pm \sqrt{4 - 4(a^2 - 2)}}{2} \]
\[ y = \frac{2 \pm \sqrt{8 - 4a^2}}{2} \]
\[ y = 1 \pm \sqrt{2 - a^2} \]

### Step 5: Determine conditions for real solutions
For \( y \) to have at least one real solution, the expression under the square root must be non-negative:
\[ 2 - a^2 \geq 0 \]
This implies:
\[ a^2 \leq 2 \]
Taking the square root gives:
\[ |a| \leq \sqrt{2} \]

### Conclusion
Thus, the equation \( \tan^4 x - 2 \sec^2 x + a^2 = 0 \) will have at least one solution if:
\[ |a| \leq \sqrt{2} \]