If `tanB= (nsinA cosA)/(1-ncos^2A)`then `tan(A +B)` equals

Options

A. `(sinA)/((1-n)cosA)`
B. `((n-1)cosA)/(sinA)`
C. `(sinA)/((n-1)cosA)`
D. `(sinA)/((n+1)cosA)`

Answer

To solve the problem, we need to find \( \tan(A + B) \) given that \( \tan B = \frac{n \sin A \cos A}{1 - n \cos^2 A} \).

### Step-by-Step Solution:

1. **Recall the formula for \( \tan(A + B) \)**:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

2. **Substitute \( \tan A \)**:
We know that \( \tan A = \frac{\sin A}{\cos A} \). Substitute this into the formula:
\[ \tan(A + B) = \frac{\frac{\sin A}{\cos A} + \tan B}{1 - \frac{\sin A}{\cos A} \tan B} \]

3. **Substitute \( \tan B \)**:
Now substitute \( \tan B = \frac{n \sin A \cos A}{1 - n \cos^2 A} \):
\[ \tan(A + B) = \frac{\frac{\sin A}{\cos A} + \frac{n \sin A \cos A}{1 - n \cos^2 A}}{1 - \frac{\sin A}{\cos A} \cdot \frac{n \sin A \cos A}{1 - n \cos^2 A}} \]

4. **Simplify the numerator**:
The numerator becomes:
\[ \frac{\sin A (1 - n \cos^2 A) + n \sin A \cos^2 A}{\cos A (1 - n \cos^2 A)} = \frac{\sin A (1 - n \cos^2 A + n \cos^2 A)}{\cos A (1 - n \cos^2 A)} = \frac{\sin A}{\cos A (1 - n \cos^2 A)} \]

5. **Simplify the denominator**:
The denominator becomes:
\[ 1 - \frac{n \sin^2 A}{1 - n \cos^2 A} = \frac{(1 - n \cos^2 A) - n \sin^2 A}{1 - n \cos^2 A} = \frac{1 - n (\sin^2 A + \cos^2 A)}{1 - n \cos^2 A} = \frac{1 - n}{1 - n \cos^2 A} \]

6. **Combine the results**:
Now we can combine the simplified numerator and denominator:
\[ \tan(A + B) = \frac{\frac{\sin A}{\cos A (1 - n \cos^2 A)}}{\frac{1 - n}{1 - n \cos^2 A}} = \frac{\sin A}{\cos A} \cdot \frac{1 - n \cos^2 A}{1 - n} = \frac{\tan A (1 - n \cos^2 A)}{1 - n} \]

7. **Final Result**:
Thus, we have:
\[ \tan(A + B) = \frac{\tan A (1 - n \cos^2 A)}{1 - n} \]