If `tan A= (1- cos B)/(sin B),` then `tan 2A` is equal to

Options

A. `tan B`
B. `tan ^(2)B`
C. `tan ^(2) B+2 tan B`
D. `tan B+2 tan B`

Answer

To find \( \tan 2A \) given that \( \tan A = \frac{1 - \cos B}{\sin B} \), we can follow these steps:

### Step 1: Express \( \tan A \) in terms of sine and cosine
We know that:
\[ \tan A = \frac{\sin A}{\cos A} \]
Given \( \tan A = \frac{1 - \cos B}{\sin B} \), we can express \( \sin A \) and \( \cos A \) in terms of \( B \).

### Step 2: Find \( \sin A \) and \( \cos A \)
Using the identity \( \tan A = \frac{\sin A}{\cos A} \), we can write:
\[ \sin A = \tan A \cdot \cos A \]
Let’s denote \( \cos A = x \). Then:
\[ \sin A = \frac{1 - \cos B}{\sin B} \cdot x \]
Using the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \):
\[ \left(\frac{1 - \cos B}{\sin B} \cdot x\right)^2 + x^2 = 1 \]

### Step 3: Substitute and simplify
Substituting \( \sin A \):
\[ \frac{(1 - \cos B)^2}{\sin^2 B} \cdot x^2 + x^2 = 1 \]
Factoring out \( x^2 \):
\[ x^2 \left(\frac{(1 - \cos B)^2}{\sin^2 B} + 1\right) = 1 \]
This gives:
\[ x^2 = \frac{1}{\frac{(1 - \cos B)^2}{\sin^2 B} + 1} \]

### Step 4: Find \( \tan 2A \)
Using the double angle formula for tangent:
\[ \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \]
Substituting \( \tan A = \frac{1 - \cos B}{\sin B} \):
\[ \tan 2A = \frac{2 \cdot \frac{1 - \cos B}{\sin B}}{1 - \left(\frac{1 - \cos B}{\sin B}\right)^2} \]

### Step 5: Simplify the expression
Calculating \( 1 - \tan^2 A \):
\[ 1 - \tan^2 A = 1 - \frac{(1 - \cos B)^2}{\sin^2 B} = \frac{\sin^2 B - (1 - \cos B)^2}{\sin^2 B} \]
Now, simplify \( \sin^2 B - (1 - \cos B)^2 \):
\[ \sin^2 B - (1 - 2\cos B + \cos^2 B) = \sin^2 B - (1 - 2\cos B + \cos^2 B) = 2\cos B - 1 \]
Thus:
\[ \tan 2A = \frac{2 \cdot \frac{1 - \cos B}{\sin B}}{\frac{2\cos B - 1}{\sin^2 B}} = \frac{2(1 - \cos B) \cdot \sin^2 B}{\sin B(2\cos B - 1)} \]
This simplifies to:
\[ \tan 2A = \frac{2(1 - \cos B) \sin B}{2\cos B - 1} \]

### Final Result
Thus, we have:
\[ \tan 2A = \frac{2(1 - \cos B) \sin B}{2\cos B - 1} \]