Certainly! Let's tackle the expression step by step:
We start with the given expression:
[tex]\[ E = \frac{\sin x}{1 - \cos x} + \frac{1 - \cos x}{\sin x}. \][/tex]
First, we work to simplify this expression.
### Step 1: Combine the fractions into a single fraction
To combine the terms into a single fraction, we get a common denominator, which is [tex]\((1 - \cos x)\sin x\)[/tex]. Thus:
[tex]\[ E = \frac{\sin x}{1 - \cos x} + \frac{1 - \cos x}{\sin x} = \frac{\sin^2 x + (1 - \cos x)^2}{(1 - \cos x)\sin x}. \][/tex]
### Step 2: Expand the numerator
Expand the numerator [tex]\(\sin^2 x + (1 - \cos x)^2\)[/tex]:
[tex]\[ (1 - \cos x)^2 = 1 - 2\cos x + \cos^2 x. \][/tex]
So,
[tex]\[ \sin^2 x + (1 - \cos x)^2 = \sin^2 x + 1 - 2\cos x + \cos^2 x. \][/tex]
### Step 3: Use Pythagorean Identity
Using the identity [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex], we can substitute [tex]\(\sin^2 x + \cos^2 x\)[/tex] with 1:
[tex]\[ \sin^2 x + 1 - 2\cos x + \cos^2 x = 1 + 1 - 2\cos x = 2 - 2\cos x = 2(1 - \cos x). \][/tex]
### Step 4: Simplify the fraction
Now the expression [tex]\(E\)[/tex] can be written as:
[tex]\[ E = \frac{2(1 - \cos x)}{(1 - \cos x)\sin x}. \][/tex]
Cancel [tex]\(1 - \cos x\)[/tex] from the numerator and the denominator (since [tex]\(x \neq n\pi\)[/tex], [tex]\(1 - \cos x\)[/tex] is not zero):
[tex]\[ E = \frac{2}{\sin x}. \][/tex]
### Step 5: Express in terms of [tex]\(\csc x\)[/tex]
Recall that [tex]\(\csc x = \frac{1}{\sin x}\)[/tex]:
[tex]\[ E = 2 \csc x. \][/tex]
### Step 6: Relate [tex]\(E\)[/tex] to the given expression
From the above, we can see that:
[tex]\[ \frac{\sin x}{1 - \cos x} + \frac{1 - \cos x}{\sin x} = 2 \csc x. \][/tex]
So, we have shown that:
[tex]\[ \frac{\sin x}{1 - \cos x} + \frac{1 - \cos x}{\sin x} \equiv k \csc x \][/tex]
where [tex]\(k = 2\)[/tex].
