If cot ` alpha = (2ab)/( a^(2) - b^(2))` then what is the value of cos `alpha ` ?

Options

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

Answer

To find the value of \( \cos \alpha \) given that \( \cot \alpha = \frac{2ab}{a^2 - b^2} \), we can follow these steps:

### Step 1: Understand the relationship between cotangent and cosine
We know that:
\[ \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \]
This means we can express \( \cos \alpha \) in terms of \( \sin \alpha \) and \( \cot \alpha \):
\[ \cos \alpha = \cot \alpha \cdot \sin \alpha \]

### Step 2: Express \( \sin \alpha \) in terms of \( \cot \alpha \)
From the cotangent definition:
\[ \sin \alpha = \frac{1}{\sqrt{1 + \cot^2 \alpha}} \]
Substituting \( \cot \alpha \):
\[ \sin \alpha = \frac{1}{\sqrt{1 + \left(\frac{2ab}{a^2 - b^2}\right)^2}} \]

### Step 3: Calculate \( \cot^2 \alpha \)
Calculating \( \cot^2 \alpha \):
\[ \cot^2 \alpha = \left(\frac{2ab}{a^2 - b^2}\right)^2 = \frac{4a^2b^2}{(a^2 - b^2)^2} \]

### Step 4: Substitute \( \cot^2 \alpha \) into the sine equation
Now substituting back into the sine equation:
\[ \sin \alpha = \frac{1}{\sqrt{1 + \frac{4a^2b^2}{(a^2 - b^2)^2}}} \]
This simplifies to:
\[ \sin \alpha = \frac{(a^2 - b^2)}{\sqrt{(a^2 - b^2)^2 + 4a^2b^2}} = \frac{(a^2 - b^2)}{\sqrt{a^4 - 2a^2b^2 + b^4 + 4a^2b^2}} = \frac{(a^2 - b^2)}{\sqrt{a^4 + 2a^2b^2 + b^4}} = \frac{(a^2 - b^2)}{(a^2 + b^2)} \]

### Step 5: Substitute \( \sin \alpha \) back to find \( \cos \alpha \)
Now substituting \( \sin \alpha \) back into the equation for \( \cos \alpha \):
\[ \cos \alpha = \cot \alpha \cdot \sin \alpha = \frac{2ab}{a^2 - b^2} \cdot \frac{(a^2 - b^2)}{(a^2 + b^2)} = \frac{2ab}{(a^2 + b^2)} \]

### Final Answer
Thus, the value of \( \cos \alpha \) is:
\[ \cos \alpha = \frac{2ab}{a^2 + b^2} \]

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