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

Mar 16, 2018

Please refer to a Proof in the Explanation.

#### Explanation:

We have, $\tan x + \tan y = \sin \frac{x}{\cos} x + \sin \frac{y}{\cos} y$,

$= \frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y}$,

$\Rightarrow \tan x + \tan y = \sin \frac{x + y}{\cos x \cos y} \ldots \ldots \ldots \ldots \ldots \ldots . \left({\square}^{1}\right)$.

Similarly, $\tan x - \tan y = \sin \frac{x - y}{\cos x \cos y} \ldots \ldots \ldots \ldots \left({\square}^{2}\right)$.

$\therefore \text{ by } \left({\square}^{1}\right) \mathmr{and} \left({\square}^{2}\right) , \frac{\tan x + \tan y}{\tan x - \tan y}$,

$= \sin \frac{x + y}{\sin} \left(x - y\right)$.

Hence, the Proof.

Mar 16, 2018

We seek to prove that:

$\sin \frac{x + y}{\sin} \left(x - y\right) \equiv \frac{\tan x + \tan y}{\tan x - \tan y}$

We can the trigonometric identities:

$\sin \left(A + B\right) \equiv \sin A \cos B + \cos A \sin B$
$\sin \left(A - B\right) \equiv \sin A \cos B - \cos A \sin B$

Consider the LHS:

$L H S = \sin \frac{x + y}{\sin} \left(x - y\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\sin x \cos y + \cos x \sin y}{\sin x \cos y - \cos x \sin y}$

Now if we multiply both numerator and denominator by $\frac{1}{\cos x \cos y}$ we get:

$L H S = \frac{\sin x \cos y + \cos x \sin y}{\sin x \cos y - \cos x \sin y} \cdot \frac{\frac{1}{\cos x \cos y}}{\frac{1}{\cos x \cos y}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\frac{\sin x \cos y}{\cos x \cos y} + \frac{\cos x \sin y}{\cos x \cos y}}{\frac{\sin x \cos y}{\cos x \cos y} - \frac{\cos x \sin y}{\cos x \cos y}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}}{\frac{\sin x}{\cos x} - \frac{\sin y}{\cos y}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\tan x + \tan y}{\tan x - \tan y} \setminus \setminus \setminus$ QED