Find `cot^(-1) ( sqrt((1-x^(2))/(1 + x^(2))))` in terms of cos

Options

A. `cos^(-1) (x^(2))`
B. `pi/2 - 1/2 cos^(-1) ( x^(2))`
C. `pi/3 - 1/2 cos^(-1) ( x^(2))`
D. None of these

Answer

To solve the problem of finding \( \cot^{-1} \left( \sqrt{\frac{1 - x^2}{1 + x^2}} \right) \) in terms of cosine, we can follow these steps:

### Step 1: Set the expression equal to an angle
Let
\[ \theta = \cot^{-1} \left( \sqrt{\frac{1 - x^2}{1 + x^2}} \right) \]
This implies that
\[ \cot \theta = \sqrt{\frac{1 - x^2}{1 + x^2}} \]

### Step 2: Express cotangent in terms of sine and cosine
Recall that
\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \]
Thus, we can write:
\[ \frac{\cos \theta}{\sin \theta} = \sqrt{\frac{1 - x^2}{1 + x^2}} \]

### Step 3: Set up a right triangle
Let’s consider a right triangle where:
- The adjacent side (base) is \( \cos \theta \)
- The opposite side (perpendicular) is \( \sin \theta \)

From the cotangent definition, we can denote:
- \( \cos \theta = \sqrt{1 - x^2} \)
- \( \sin \theta = \sqrt{1 + x^2} \)

### Step 4: Find the hypotenuse
Using the Pythagorean theorem, the hypotenuse \( h \) can be calculated as:
\[ h = \sqrt{(\cos \theta)^2 + (\sin \theta)^2} = \sqrt{(\sqrt{1 - x^2})^2 + (\sqrt{1 + x^2})^2} \]
This simplifies to:
\[ h = \sqrt{(1 - x^2) + (1 + x^2)} = \sqrt{2} \]

### Step 5: Find cosine
Now we can find \( \cos \theta \):
\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{1 - x^2}}{\sqrt{2}} = \frac{1 - x^2}{\sqrt{2}} \]

### Step 6: Express \( \theta \) in terms of cosine
Thus, we can express \( \theta \) as:
\[ \theta = \cos^{-1} \left( \frac{\sqrt{1 - x^2}}{\sqrt{2}} \right) \]

### Final Result
Therefore, we have:
\[ \cot^{-1} \left( \sqrt{\frac{1 - x^2}{1 + x^2}} \right) = \cos^{-1} \left( \frac{\sqrt{1 - x^2}}{\sqrt{2}} \right) \]