# Question #ba262

Jul 6, 2016

The proof is a bit long, but manageable. See below.

#### Explanation:

When trying to prove trig identities involving fractions, it's always a good idea to add the fractions first:
$\sin \frac{t}{1 - \cos t} + \frac{1 + \cos t}{\sin} t = \frac{2 \left(1 + \cos t\right)}{\sin} t$

$\to \sin \frac{t}{1 - \cos t} \sin \frac{t}{\sin} t + \frac{1 + \cos t}{\sin} t \frac{1 - \cos t}{1 - \cos t} = \frac{2 \left(1 + \cos t\right)}{\sin} t$

$\to {\sin}^{2} \frac{t}{\left(1 - \cos t\right) \left(\sin t\right)} + \frac{\left(1 + \cos t\right) \left(1 - \cos t\right)}{\left(1 - \cos t\right) \left(\sin t\right)} = \frac{2 \left(1 + \cos t\right)}{\sin} t$

$\to \frac{{\sin}^{2} t + \left(1 + \cos t\right) \left(1 - \cos t\right)}{\left(1 - \cos t\right) \left(\sin t\right)} = \frac{2 \left(1 + \cos t\right)}{\sin} t$

The expression $\left(1 + \cos t\right) \left(1 - \cos t\right)$ is actually a difference of squares in disguise:
$\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$
With $a = 1$ and $b = \cos t$. It evaluates to ${\left(1\right)}^{2} - {\left(\cos t\right)}^{2} = 1 - {\cos}^{2} t$.

We can go even further with $1 - {\cos}^{2} t$. Recall the basic Pythagorean Identity:
${\cos}^{2} x + {\sin}^{2} x = 1$

Subtracting ${\cos}^{2} x$ from both sides, we see:
${\sin}^{2} x = 1 - {\cos}^{2} x$

Since $x$ is just a placeholder variable, we can say that ${\sin}^{2} t = 1 - {\cos}^{2} t$. Therefore, the $\left(1 + \cos t\right) \left(1 - \cos t\right)$ becomes ${\sin}^{2} t$:
$\frac{{\sin}^{2} t + {\sin}^{2} t}{\left(1 - \cos t\right) \left(\sin t\right)} = \frac{2 \left(1 + \cos t\right)}{\sin} t$
$\to \frac{2 {\sin}^{2} t}{\left(1 - \cos t\right) \left(\sin t\right)} = \frac{2 \left(1 + \cos t\right)}{\sin} t$

Note that sines cancel:
$\frac{2 {\cancel{{\sin}^{2} t}}^{\sin} t}{\left(1 - \cos t\right) \cancel{\left(\sin t\right)}} = \frac{2 \left(1 + \cos t\right)}{\sin} t$
$\to \frac{2 \sin t}{1 - \cos t} = \frac{2 \left(1 + \cos t\right)}{\sin} t$

We're almost done. The last step is to multiply the left side by the conjugate of $1 - \cos t$ (which is $1 + \cos t$), to take advantage of the difference of squares property:
$\frac{2 \sin t}{1 - \cos t} \frac{1 + \cos t}{1 + \cos t} = \frac{2 \left(1 + \cos t\right)}{\sin} t$

$\to \frac{2 \sin t \left(1 + \cos t\right)}{\left(1 - \cos t\right) \left(1 + \cos t\right)} = \frac{2 \left(1 + \cos t\right)}{\sin} t$

Again, we can see that $\left(1 - \cos t\right) \left(1 + \cos t\right)$ is a difference of squares, with $a = 1$ and $b = \cos t$. It evaluates to ${\left(1\right)}^{2} - {\left(\cos t\right)}^{2}$, or $1 - {\cos}^{2} t$. We already showed that ${\sin}^{2} t = 1 - {\cos}^{2} t$, so the denominator gets replaced:
$\frac{2 \sin t \left(1 + \cos t\right)}{{\sin}^{2} t} = \frac{2 \left(1 + \cos t\right)}{\sin} t$

Sines cancel:
$\frac{2 \cancel{\sin t} \left(1 + \cos t\right)}{{\cancel{{\sin}^{2} t}}^{\sin} t} = \frac{2 \left(1 + \cos t\right)}{\sin} t$

And voila, proof complete:
$\frac{2 \left(1 + \cos t\right)}{\sin} t = \frac{2 \left(1 + \cos t\right)}{\sin} t$

Jul 6, 2016

Let me try

#### Explanation:

$L H S = \sin \frac{t}{1 - \cos t} + \frac{1 + \cos t}{\sin} t$

Inspecting the RHS we take common$\frac{1 + \cos t}{\sin} t$

So
$L H S = \frac{1 + \cos t}{\sin} t \left(\sin \frac{t}{1 + \cos t} \cdot \sin \frac{t}{1 - \cos t} + 1\right)$

$= \frac{1 + \cos t}{\sin} t \left({\sin}^{2} \frac{t}{1 - {\cos}^{2} t} + 1\right)$

$= \frac{1 + \cos t}{\sin} t \left({\sin}^{2} \frac{t}{\sin} ^ 2 t + 1\right)$

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

$= \frac{2 \left(1 + \cos t\right)}{\sin} t = R H S$

Proved