The equation `tan^4x-2sec^2x+a=0` will have at least one solution if `1

Options

A. `1 lt a le 4`
B. `a ge 2`
C. `a le 3`
D. none of these

Explanation

To solve the equation \( \tan^4 x - 2 \sec^2 x + a = 0 \) for the condition that it has at least one solution, we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \tan^4 x - 2 \sec^2 x + a = 0 \]
Using the identity \( \sec^2 x = 1 + \tan^2 x \), we can rewrite \( \sec^2 x \) in terms of \( \tan^2 x \):
\[ \tan^4 x - 2(1 + \tan^2 x) + a = 0 \]
This simplifies to:
\[ \tan^4 x - 2 - 2\tan^2 x + a = 0 \]
Rearranging gives:
\[ \tan^4 x - 2\tan^2 x + (a - 2) = 0 \]

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

### Step 3: Analyze the quadratic equation
For the quadratic equation \( y^2 - 2y + (a - 2) = 0 \) to have at least one solution, the discriminant must be non-negative:
\[ D = b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot (a - 2) \geq 0 \]
Calculating the discriminant:
\[ 4 - 4(a - 2) \geq 0 \]
This simplifies to:
\[ 4 - 4a + 8 \geq 0 \]
\[ 12 - 4a \geq 0 \]
\[ 4a \leq 12 \]
\[ a \leq 3 \]

### Step 4: Conclusion
Thus, for the equation to have at least one solution, \( a \) must satisfy:
\[ a \leq 3 \]
From the given options, the correct answer is:
(c) \( a \leq 3 \)