# Question #5bf52

Apr 8, 2017

You have to use some trigonometric formulae.

#### Explanation:

I assume that what you mean by "sen" is actually "sin".

Let's see. So you have this on the left-hand-side:
$\frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y - \sin x \sin y}$

This obviously simplifies to the following:
$\sin \frac{x + y}{\cos} \left(x + y\right)$
because you use the "sicosico" and "cocosisi" formulae.
Don't know them? It's easy. All you need to remember is that
1. sine and cosine is applied once on x and y, and
2. "sicosico" keeps the sign, while "cocosisi" inverts the sign.
What sign? you may ask. Well, it is the sign that is inside the parentheses, of course.
See below:

$\sin \left(x + y\right) = \sin x \cos y + \sin y \cos x$
and
$\cos \left(x + y\right) = \cos x \cos y - \sin x \sin y$

Notice how the sign is a minus in the "cocosisi" case?
OK, let's move on.

We can simplify of course what we got:
$\frac{\sin \left(x + y\right)}{\cos \left(x + y\right)} = \tan \left(x + y\right)$

Now, for the right-hand-side, we have:
$\frac{\tan x + \tan y}{1 - \tan x \tan y}$

The way to deal with tangeant is to convert them back to sine and cosine. $\tan x = \sin \frac{x}{\cos} x$.
We now have:
$\frac{\tan x + \tan y}{1 - \tan x \tan y} = \frac{\sin \frac{x}{\cos} x + \sin \frac{y}{\cos} y}{1 - \frac{\sin x \cdot \sin y}{\cos x \cdot \cos y}}$
which becomes
$= \frac{\frac{\sin x \cos y + \sin y \cos x}{\cos x \cos y}}{\frac{\cos x \cos y - \sin x \sin y}{\cos x \cos y}}$

Now, cancel out the $\frac{1}{\cos x \cos y}$ from the numerator and denominator, we have:

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

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

The left-hand-side is the same as the right-hand-side.
Q.E.D.