Let `f(x) = cos(tan^(-1) ( sin ( cot^(-1)x)))`. The simplest form of f(x) can be written as `((x^(2) +A)/(x^(2) +B))^(1//2)` . Then the value of `(A + B) ` is ………………

Answer

To solve the problem step by step, we will simplify the function \( f(x) = \cos(\tan^{-1}(\sin(\cot^{-1}(x)))) \).

### Step 1: Rewrite \( \cot^{-1}(x) \)
Let \( \theta = \cot^{-1}(x) \). By definition, this implies that:
\[ x = \cot(\theta) = \frac{\text{adjacent}}{\text{opposite}} = \frac{x}{1} \]
We can visualize this in a right triangle where the adjacent side is \( x \) and the opposite side is \( 1 \).

### Step 2: Find the hypotenuse
Using the Pythagorean theorem, we can find the hypotenuse \( h \):
\[ h = \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1} \]

### Step 3: Find \( \sin(\theta) \)
From the triangle, we can find \( \sin(\theta) \):
\[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 4: Substitute into \( f(x) \)
Now we substitute \( \sin(\theta) \) back into our function:
\[ f(x) = \cos(\tan^{-1}(\sin(\theta))) = \cos(\tan^{-1}\left(\frac{1}{\sqrt{x^2 + 1}}\right)) \]

### Step 5: Let \( \phi = \tan^{-1}\left(\frac{1}{\sqrt{x^2 + 1}}\right) \)
From the definition of \( \tan \):
\[ \tan(\phi) = \frac{1}{\sqrt{x^2 + 1}} \]
This means we can form another triangle where:
- Opposite side = 1
- Adjacent side = \( \sqrt{x^2 + 1} \)

### Step 6: Find the hypotenuse for \( \phi \)
Using the Pythagorean theorem again:
\[ \text{hypotenuse} = \sqrt{1^2 + (\sqrt{x^2 + 1})^2} = \sqrt{1 + (x^2 + 1)} = \sqrt{x^2 + 2} \]

### Step 7: Find \( \cos(\phi) \)
Now we can find \( \cos(\phi) \):
\[ \cos(\phi) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2 + 2}} \]

### Step 8: Final expression for \( f(x) \)
Thus, we can express \( f(x) \) as:
\[ f(x) = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2 + 2}} = \left(\frac{x^2 + 1}{x^2 + 2}\right)^{1/2} \]

### Step 9: Identify \( A \) and \( B \)
Comparing this with the form \( \left(\frac{x^2 + A}{x^2 + B}\right)^{1/2} \), we can see:
- \( A = 1 \)
- \( B = 2 \)

### Step 10: Calculate \( A + B \)
Finally, we find:
\[ A + B = 1 + 2 = 3 \]

Thus, the value of \( A + B \) is \( \boxed{3} \).

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