solve the following equation `sec^(-1).(x)/(a) - sec^(-1).(x)/(b) = sec^(-1) b - sec^(-1) a, a ge 1, b ge 1, a!= b`

Explanation

To solve the equation

\[ \sec^{-1}\left(\frac{x}{a}\right) - \sec^{-1}\left(\frac{x}{b}\right) = \sec^{-1}(b) - \sec^{-1}(a) \]

where \( a \geq 1 \), \( b \geq 1 \), and \( a \neq b \), we will follow these steps:

### Step 1: Rewrite the equation using cosine
We know that \(\sec^{-1}(y) = \cos^{-1}\left(\frac{1}{y}\right)\). Therefore, we can rewrite the equation as:

\[ \cos^{-1}\left(\frac{a}{x}\right) - \cos^{-1}\left(\frac{b}{x}\right) = \cos^{-1}\left(\frac{1}{b}\right) - \cos^{-1}\left(\frac{1}{a}\right) \]

### Step 2: Use the cosine addition formula
Using the formula for the difference of cosines, we can express the left-hand side as:

\[ \cos^{-1}\left(\frac{a}{x}\right) + \cos^{-1}\left(\frac{1}{a}\right) = \cos^{-1}\left(\frac{b}{x}\right) + \cos^{-1}\left(\frac{1}{b}\right) \]

### Step 3: Set up the equation
Now, we can equate the two sides:

\[ \cos^{-1}\left(\frac{a}{x}\right) + \cos^{-1}\left(\frac{1}{a}\right) = \cos^{-1}\left(\frac{b}{x}\right) + \cos^{-1}\left(\frac{1}{b}\right) \]

### Step 4: Eliminate the cosines
To eliminate the cosines, we can take the cosine of both sides. This leads us to:

\[ \frac{a}{x} = \frac{b}{x} \]

### Step 5: Rearranging the equation
Rearranging gives us:

\[ \frac{a}{x} - \frac{b}{x} = 0 \]

This simplifies to:

\[ a - b = 0 \]

### Step 6: Solve for \(x\)
Since \(a \neq b\) as per the problem statement, we can conclude that:

\[ x^2 = ab \]

Taking the square root gives:

\[ x = ab \]

### Step 7: Consider the conditions
Given that \(a \geq 1\) and \(b \geq 1\), \(x\) must be positive. Thus, we have:

\[ x = ab \]

### Final Solution
The solution to the equation is:

\[ x = ab \]