If a and b are positive quantities, `(a gt b)` find minimum positive value of `(a sectheta- b tantheta)`

Options

A. 2ab
B. `sqrt(a^2-b^2)`
C. a-b
D. `sqrt(a^2+b^2)`

Explanation

To find the minimum positive value of \( S = a \sec \theta - b \tan \theta \) given that \( a \) and \( b \) are positive quantities and \( a > b \), we can follow these steps:

### Step 1: Define the expression
Let \( S = a \sec \theta - b \tan \theta \).

### Step 2: Rearrange the equation
Rearranging gives us:
\[ S + b \tan \theta = a \sec \theta \]

### Step 3: Square both sides
Squaring both sides results in:
\[ (S + b \tan \theta)^2 = (a \sec \theta)^2 \]
This expands to:
\[ S^2 + 2bS \tan \theta + b^2 \tan^2 \theta = a^2 \sec^2 \theta \]

### Step 4: Use the identity for secant
Recall the trigonometric identity \( \sec^2 \theta = 1 + \tan^2 \theta \). Thus, we can write:
\[ a^2 \sec^2 \theta = a^2 (1 + \tan^2 \theta) = a^2 + a^2 \tan^2 \theta \]

### Step 5: Substitute back into the equation
Substituting this back gives:
\[ S^2 + 2bS \tan \theta + b^2 \tan^2 \theta = a^2 + a^2 \tan^2 \theta \]

### Step 6: Rearranging the equation
Rearranging leads to:
\[ b^2 \tan^2 \theta - a^2 \tan^2 \theta + 2bS \tan \theta + S^2 + a^2 = 0 \]
This can be rewritten as:
\[ (b^2 - a^2) \tan^2 \theta + 2bS \tan \theta + (S^2 + a^2) = 0 \]

### Step 7: Identify as a quadratic in \( \tan \theta \)
This is a quadratic equation in \( \tan \theta \):
\[ A \tan^2 \theta + B \tan \theta + C = 0 \]
where \( A = b^2 - a^2 \), \( B = 2bS \), and \( C = S^2 + a^2 \).

### Step 8: Apply the discriminant condition
For the quadratic to have real roots, the discriminant must be non-negative:
\[ B^2 - 4AC \geq 0 \]
Substituting the values gives:
\[ (2bS)^2 - 4(b^2 - a^2)(S^2 + a^2) \geq 0 \]

### Step 9: Simplifying the discriminant
Expanding this leads to:
\[ 4b^2S^2 - 4(b^2 - a^2)(S^2 + a^2) \geq 0 \]
This simplifies to:
\[ 4b^2S^2 - 4b^2S^2 + 4a^2(b^2 - a^2) \geq 0 \]
which simplifies to:
\[ 4a^2(b^2 - a^2) \geq 0 \]

### Step 10: Finding the minimum positive value
Since \( a > b \), we have \( b^2 - a^2 < 0 \). Thus, we can conclude:
\[ S^2 \geq a^2 - b^2 \]
Taking the square root gives:
\[ S \geq \sqrt{a^2 - b^2} \]

### Conclusion
The minimum positive value of \( S = a \sec \theta - b \tan \theta \) is:
\[ \sqrt{a^2 - b^2} \]