# How do you prove  {(sin(x + y)) / sin(x)cos(y)} = 1-cot(x)tan(y)?

Apr 7, 2017

Q.E.D. See below.

#### Explanation:

I'm going to assume you meant $\sin \frac{x + y}{\sin x \cos y} = 1 + \cot x \tan y$ and just forgot parentheses on the left side denominator and the plus on the right side, otherwise the identity is false.

Let's start working the left hand side by using the sine addition formula $\sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$
$\sin \frac{x + y}{\sin x \cos y} = \frac{\sin x \cos y + \cos x \sin y}{\sin x \cos y}$

Then split the numerator over the denominator
$\frac{\sin x \cos y + \cos x \sin y}{\sin x \cos y} = \frac{\sin x \cos y}{\sin x \cos y} + \frac{\cos x \sin y}{\sin x \cos y}$

Simplify and factor
$\frac{\sin x \cos y}{\sin x \cos y} + \frac{\cos x \sin y}{\sin x \cos y} = 1 + \left(\cos \frac{x}{\sin} x\right) \left(\sin \frac{y}{\cos} y\right)$

And knowing that $\tan \theta = \sin \frac{\theta}{\cos} \theta$ and $\cot \theta = \cos \frac{\theta}{\sin} \theta$,
$1 + \left(\cos \frac{x}{\sin} x\right) \left(\sin \frac{y}{\cos} y\right) = 1 + \cot x \tan y$

Therefore the left side is equal to the right side. Q.E.D.

Apr 7, 2017

Assuming the questioner meant $\frac{\sin \left(x + y\right)}{\sin x \cos y} = 1 + \cot x \tan y$, then the proof is as follows:

#### Explanation:

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

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

$1 + \cot x \tan y$ QED/ΟΕΔ