If `costheta=(x^(2)-y^(2))/(x^(2)+y^(2))` then the value of `cottheta` is equal to (If `0lethetale90^(@))`

Options

A. `(2xy)/(x^(2)-y^(2))`
B. `(2xy)/(x^(2)+y^(2))`
C. `(x^(2)+y^(2))/(2xy)`
D. `(x^(2)-y^(2))/(2xy)`

Answer

To find the value of \( \cot \theta \) given that \( \cos \theta = \frac{x^2 - y^2}{x^2 + y^2} \), we can follow these steps:

### Step 1: Understand the relationship between trigonometric functions
We know that:
\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \]
We need to express \( \sin \theta \) in terms of \( \cos \theta \).

### Step 2: Use the Pythagorean identity
According to the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
We can express \( \sin^2 \theta \) as:
\[ \sin^2 \theta = 1 - \cos^2 \theta \]

### Step 3: Substitute \( \cos \theta \)
Substituting \( \cos \theta \) into the identity:
\[ \sin^2 \theta = 1 - \left(\frac{x^2 - y^2}{x^2 + y^2}\right)^2 \]

### Step 4: Simplify \( \sin^2 \theta \)
Calculating \( \left(\frac{x^2 - y^2}{x^2 + y^2}\right)^2 \):
\[ \sin^2 \theta = 1 - \frac{(x^2 - y^2)^2}{(x^2 + y^2)^2} \]
Now, we can find a common denominator:
\[ \sin^2 \theta = \frac{(x^2 + y^2)^2 - (x^2 - y^2)^2}{(x^2 + y^2)^2} \]

### Step 5: Expand and simplify the numerator
Expanding the numerator:
\[ (x^2 + y^2)^2 - (x^2 - y^2)^2 = (x^4 + 2x^2y^2 + y^4) - (x^4 - 2x^2y^2 + y^4) = 4x^2y^2 \]
Thus, we have:
\[ \sin^2 \theta = \frac{4x^2y^2}{(x^2 + y^2)^2} \]

### Step 6: Find \( \sin \theta \)
Taking the square root gives us:
\[ \sin \theta = \frac{2xy}{x^2 + y^2} \]

### Step 7: Calculate \( \cot \theta \)
Now substituting back into the formula for \( \cot \theta \):
\[ \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{x^2 - y^2}{x^2 + y^2}}{\frac{2xy}{x^2 + y^2}} = \frac{x^2 - y^2}{2xy} \]

### Final Result
Thus, the value of \( \cot \theta \) is:
\[ \cot \theta = \frac{x^2 - y^2}{2xy} \]