First slide

Trigonometric transformations

Question

$\frac{\sin (A + B)}{\sin (A - B)} = \frac{x + y}{x - y}$ then $\frac{Tan A}{Tan B} =$

Easy

A

$\frac{x}{y}$
B

$\frac{y}{x}$
C

$\frac{x}{2 y}$
D

$\frac{2 x}{y}$

Solution

$\frac{\sin (A + B)}{\sin (A - B)} = \frac{x + y}{x - y}$ $\Rightarrow$ by componendo & dividendo
$\frac{\sin (A + B) + \sin (A - B)}{\sin (A + B) - \sin (A - B)} = \frac{x + y + x - y}{x + y - x + y}$
$\frac{2 \sin A \cos B}{2 \cos A \sin B} = \frac{x}{y} \Rightarrow \frac{\tan A}{\tan B} = \frac{x}{y}$

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