If `tan A=(1-cosB)/(sinB)`, then tan 2A is equal to

Options

A. cot B
B. tan B
C. cos B
D. cosec B

Answer

To solve the problem where \( \tan A = \frac{1 - \cos B}{\sin B} \) and we need to find \( \tan 2A \), we can follow these steps:

### Step 1: Use the Double Angle Formula for Tangent
The double angle formula for tangent is given by:
\[ \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \]

### Step 2: Substitute the Given Value of \( \tan A \)
We know that:
\[ \tan A = \frac{1 - \cos B}{\sin B} \]
Now, we will substitute this into the double angle formula:
\[ \tan 2A = \frac{2 \left(\frac{1 - \cos B}{\sin B}\right)}{1 - \left(\frac{1 - \cos B}{\sin B}\right)^2} \]

### Step 3: Simplify the Denominator
First, we need to simplify the denominator:
\[ 1 - \left(\frac{1 - \cos B}{\sin B}\right)^2 = 1 - \frac{(1 - \cos B)^2}{\sin^2 B} \]
Using the identity \( \sin^2 B = 1 - \cos^2 B \), we can rewrite this as:
\[ 1 - \frac{(1 - \cos B)^2}{1 - \cos^2 B} \]

### Step 4: Find a Common Denominator
To simplify further, we can find a common denominator:
\[ = \frac{(1 - \cos^2 B) - (1 - \cos B)^2}{1 - \cos^2 B} \]
Expanding the numerator:
\[ = \frac{(1 - \cos^2 B) - (1 - 2\cos B + \cos^2 B)}{1 - \cos^2 B} \]
This simplifies to:
\[ = \frac{2\cos B}{1 - \cos^2 B} \]

### Step 5: Substitute Back into the Formula
Now substituting back into the formula for \( \tan 2A \):
\[ \tan 2A = \frac{2 \left(\frac{1 - \cos B}{\sin B}\right)}{\frac{2\cos B}{1 - \cos^2 B}} \]
This simplifies to:
\[ = \frac{(1 - \cos B)(1 - \cos^2 B)}{\sin B \cos B} \]

### Step 6: Final Simplification
Using \( 1 - \cos^2 B = \sin^2 B \):
\[ = \frac{(1 - \cos B)\sin^2 B}{\sin B \cos B} \]
This simplifies to:
\[ = \frac{(1 - \cos B)\sin B}{\cos B} \]
Thus, we have:
\[ \tan 2A = \frac{(1 - \cos B)\sin B}{\cos B} \]

### Conclusion
Thus, the final answer for \( \tan 2A \) in terms of \( B \) is:
\[ \tan 2A = \frac{(1 - \cos B)\sin B}{\cos B} \]