# Question 38190

Aug 14, 2017

See below.

#### Explanation:

Rule number two or three for identities (depending on the math teacher): when you see fractions, add them. It's an extremely useful step that reveals more information than what you have to begin with.

Our common denominator for the addition here is $\left(\sin\right) \left(1 + \cos\right)$. For the first term:
$\frac{\sin}{1 + \cos}$

If we want $\left(\sin\right) \left(1 + \cos\right)$ on the bottom, then we need to multiply the bottom by $\sin$; and what we do to the bottom, we do to the top. Therefore, this becomes:
$\frac{\left(\sin\right) \left(\sin\right)}{\left(1 + \cos\right) \left(\sin\right)}$
$= \frac{\left(\sin\right) \left(\sin\right)}{\left(1 + \cos\right) \left(\sin\right)}$

Note that I'm not in a hurry to simplify anything further. Leave everything by itself (for instance, don't combine $\left(1 + \cos\right) \left(\sin\right)$ to make $\sin + \left(\sin\right) \left(\cos\right)$ because this makes it more complicated than it needs to be).

Likewise, we multiply the second term by $\frac{1 + \cos}{1 + \cos}$, to get:
$\frac{\left(1 + \cos\right) \left(1 + \cos\right)}{\left(\sin\right) \left(1 + \cos\right)}$

We now have:
$\frac{\left(\sin\right) \left(\sin\right)}{\left(1 + \cos\right) \left(\sin\right)} + \frac{\left(1 + \cos\right) \left(1 + \cos\right)}{\left(\sin\right) \left(1 + \cos\right)}$

$\frac{\left(\sin\right) \left(\sin\right) + \left(1 + \cos\right) \left(1 + \cos\right)}{\left(\sin\right) \left(1 + \cos\right)}$

Now we can multiply everything out, since we're obviously getting nowhere with the above expression:
$\frac{{\sin}^{2} + 2 \cos + {\cos}^{2} + 1}{\left(\sin\right) \left(1 + \cos\right)}$

Recognize a Pythagorean Identity in here? I think this is the most challenging part of identities - at least, from what I saw from my classmates in precalc. It's kind of hard spotting identities in expressions like these, and it takes tons of practice. Recall that ${\sin}^{2} + {\cos}^{2} = 1$; this is probably the most important identity in trig. It's also hiding in the numerator, right here:
$\frac{\textcolor{red}{{\sin}^{2}} + 2 \cos + \textcolor{red}{{\cos}^{2}} + 1}{\left(\sin\right) \left(1 + \cos\right)}$

We can replace that with $1$, since it equals $1$:
$\frac{\textcolor{red}{1} + 2 \cos + 1}{\left(\sin\right) \left(1 + \cos\right)}$

And add the two ones in the numerator to get $2$:
$\frac{2 + 2 \cos}{\left(\sin\right) \left(1 + \cos\right)}$

Since both terms in the numerator have a common factor of two, we might try factoring out the two:
$\frac{2 \left(1 + \cos\right)}{\left(\sin\right) \left(1 + \cos\right)}$

And what do you know, the $1 + \cos$ cancels:
$\frac{2 \cancel{\left(1 + \cos\right)}}{\left(\sin\right) \cancel{\left(1 + \cos\right)}}$
=2/sin)=2csc->since 1/sin)=csc#

And boom, we're done. Also, I've been using $\sin$ and $\cos$ to make the answer easier on the eyes, but we really should be using $\sin x$ and $\cos x$. Doesn't matter to me, but it does matter to the people giving out the grades.