If `tan^(2)x+secx -a = 0` has atleast one solution, then complete set of values of a is :

Options

A. `(-oo, 1]`
B. `[-1, oo)`
C. `[(9)/(4) , oo)`
D. `[1, oo)`

Answer

To solve the equation \( \tan^2 x + \sec x - a = 0 \) and find the complete set of values of \( a \) such that the equation has at least one solution, we can follow these steps:

### Step 1: Rewrite the equation
Start with the equation:
\[ \tan^2 x + \sec x - a = 0 \]
We can rearrange it to:
\[ \tan^2 x = a - \sec x \]

### Step 2: Use the trigonometric identity
Recall the identity:
\[ 1 + \tan^2 x = \sec^2 x \]
From this, we can express \( \tan^2 x \) in terms of \( \sec x \):
\[ \tan^2 x = \sec^2 x - 1 \]
Substituting this into our equation gives:
\[ \sec^2 x - 1 = a - \sec x \]

### Step 3: Rearrange the equation
Rearranging the equation leads to:
\[ \sec^2 x + \sec x - a + 1 = 0 \]
Let \( t = \sec x \). The equation now becomes:
\[ t^2 + t - (a - 1) = 0 \]

### Step 4: Determine the condition for real solutions
For the quadratic equation \( t^2 + t - (a - 1) = 0 \) to have at least one real solution, the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Here, \( a = 1 \), \( b = 1 \), and \( c = -(a - 1) \). Thus, the discriminant is:
\[ D = 1^2 - 4 \cdot 1 \cdot (-(a - 1)) = 1 + 4(a - 1) \]
This simplifies to:
\[ D = 4a - 3 \]
Setting the discriminant greater than or equal to zero gives:
\[ 4a - 3 \geq 0 \]

### Step 5: Solve the inequality
Solving the inequality:
\[ 4a \geq 3 \implies a \geq \frac{3}{4} \]

### Step 6: Conclusion
Thus, the complete set of values of \( a \) such that the equation \( \tan^2 x + \sec x - a = 0 \) has at least one solution is:
\[ a \in \left[\frac{3}{4}, \infty\right) \]