If `sqrt((1+cosA)/(1-cosA))=(x)/(y)` then the value of tan A is

Options

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

Explanation

To solve the equation \( \sqrt{\frac{1 + \cos A}{1 - \cos A}} = \frac{x}{y} \) and find the value of \( \tan A \), we can follow these steps:

### Step 1: Use Trigonometric Identities
We start by using the trigonometric identities for \( 1 + \cos A \) and \( 1 - \cos A \):
\[ 1 + \cos A = 2 \cos^2\left(\frac{A}{2}\right) \]
\[ 1 - \cos A = 2 \sin^2\left(\frac{A}{2}\right) \]

### Step 2: Substitute the Identities
Substituting these identities into the original equation gives:
\[ \sqrt{\frac{2 \cos^2\left(\frac{A}{2}\right)}{2 \sin^2\left(\frac{A}{2}\right)}} = \frac{x}{y} \]

### Step 3: Simplify the Expression
The \( 2 \) in the numerator and denominator cancels out:
\[ \sqrt{\frac{\cos^2\left(\frac{A}{2}\right)}{\sin^2\left(\frac{A}{2}\right)}} = \frac{x}{y} \]
This simplifies to:
\[ \sqrt{\cot^2\left(\frac{A}{2}\right)} = \frac{x}{y} \]
Since \( \sqrt{\cot^2\left(\frac{A}{2}\right)} = \cot\left(\frac{A}{2}\right) \), we have:
\[ \cot\left(\frac{A}{2}\right) = \frac{x}{y} \]

### Step 4: Find the Reciprocal
Taking the reciprocal gives:
\[ \tan\left(\frac{A}{2}\right) = \frac{y}{x} \]

### Step 5: Use the Double Angle Formula for Tangent
Using the double angle formula for tangent:
\[ \tan A = \frac{2 \tan\left(\frac{A}{2}\right)}{1 - \tan^2\left(\frac{A}{2}\right)} \]
Substituting \( \tan\left(\frac{A}{2}\right) = \frac{y}{x} \):
\[ \tan A = \frac{2 \cdot \frac{y}{x}}{1 - \left(\frac{y}{x}\right)^2} \]

### Step 6: Simplify the Expression
This becomes:
\[ \tan A = \frac{\frac{2y}{x}}{1 - \frac{y^2}{x^2}} = \frac{\frac{2y}{x}}{\frac{x^2 - y^2}{x^2}} = \frac{2y \cdot x^2}{x(x^2 - y^2)} = \frac{2xy}{x^2 - y^2} \]

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