Question

If sin A = n sin(A + 2 B), then tan(A + B) =}

• A. $\frac{1+n}{2-n} \cdot \tan B$
• B. $\frac{1-n}{1+n} \cdot \tan B$
• C. $\frac{1-n}{2+n} \cdot \tan B$
• D. $\frac{1+n}{1-n} \cdot \tan B$

Hint:

Use Componendo and Dividendo when dealing with ratios of sine/cosine sums.

Answer 1

The Correct Option is D

Step 1: Rearrange
$\frac{\sin(A + 2B)}{\sin A} = \frac{1}{n}$.
Step 2: Componendo and Dividendo
$\frac{\sin(A + 2B) + \sin A}{\sin(A + 2B) - \sin A} = \frac{1 + n}{1 - n}$.
$\frac{2 \sin(A+B) \cos B}{2 \cos(A+B) \sin B} = \frac{1+n}{1-n}$.
Step 3: Simplify
$\tan(A+B) \cot B = \frac{1+n}{1-n} \implies \tan(A+B) = \frac{1+n}{1-n} \tan B$.
Final Answer: (D)