# How do you prove [sin(x+ y) - sin(x-y)] /[ cos(x+ y) + cos(x-y)]= tan y?

Jun 26, 2016

see explanation

#### Explanation:

To $\textcolor{b l u e}{\text{Prove}}$ we require to manipulate one side into the same form as the other side.This will involve using $\textcolor{b l u e}{\text{Addition formulae}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminders}}$

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Starting with the left side and simplifying numerator/denominator separately.

Numerator

$\sin x \cos y + \cos x \sin y - \left[\sin x \cos y - \cos x \sin y\right)$

$= \cancel{\sin x \cos y} + \cos x \sin y - \cancel{\sin x \cos y} + \cos x \sin y$

$= 2 \cos x \sin y$

Denominator

$\cos x \cos y - \sin x \sin y + \cos x \cos y + \sin x \sin y$

$= \cos x \cos y - \cancel{\sin x \sin y} + \cos x \cos y + \cancel{\sin x \sin y}$

$= 2 \cos x \cos y$
$\text{---------------------------------------------------------------}$

left side can now be expressed as

$\frac{2 \cos x \sin y}{2 \cos x \cos y} = \frac{\cancel{2} \cancel{\cos x} \sin y}{\cancel{2} \cancel{\cos x} \cos y} = \frac{\sin y}{\cos y}$

and $\frac{\sin y}{\cos y} = \tan y = \text{right side hence proved}$