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 can follow these steps:

### Step 1: Rewrite the equation
We start with the original equation:
\[ \tan^4 x - 2 \sec^2 x + a^2 = 0 \]
We know that \( \sec^2 x = 1 + \tan^2 x \). Therefore, we can substitute \( \sec^2 x \) in the equation:
\[ \tan^4 x - 2(1 + \tan^2 x) + a^2 = 0 \]
This simplifies to:
\[ \tan^4 x - 2 - 2\tan^2 x + a^2 = 0 \]

### Step 2: Rearranging the equation
Rearranging the equation gives us:
\[ \tan^4 x - 2\tan^2 x + (a^2 - 2) = 0 \]
Let \( y = \tan^2 x \). Then we can rewrite the equation as:
\[ y^2 - 2y + (a^2 - 2) = 0 \]

### Step 3: Finding the discriminant
For the quadratic equation \( y^2 - 2y + (a^2 - 2) = 0 \) to have at least one real solution, the discriminant must be non-negative. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac = (-2)^2 - 4(1)(a^2 - 2) \]
Calculating this gives:
\[ D = 4 - 4(a^2 - 2) = 4 - 4a^2 + 8 = 12 - 4a^2 \]

### Step 4: Setting the discriminant condition
For the quadratic to have at least one solution, we require:
\[ D \geq 0 \]
Thus, we have:
\[ 12 - 4a^2 \geq 0 \]
This simplifies to:
\[ 4a^2 \leq 12 \]
Dividing both sides by 4 gives:
\[ a^2 \leq 3 \]

### Step 5: Conclusion
Taking the square root of both sides, we find:
\[ |a| \leq \sqrt{3} \]
This means that \( a \) must satisfy:
\[ -a \leq \sqrt{3} \quad \text{and} \quad a \leq \sqrt{3} \]
Thus, the condition for the equation to have at least one solution is:
\[ |a| \leq \sqrt{3} \]

### Final Answer
The equation \( \tan^4 x - 2 \sec^2 x + a^2 = 0 \) will have at least one solution if:
\[ |a| \leq \sqrt{3} \]