Sure! Let's complete the proof step-by-step. We want to verify that [tex]\(\cot^2 \left(\frac{x}{2}\right) = \frac{\cos x + 1}{\cos x - 1}\)[/tex].
### Step 1: Recall the cotangent identity
The cotangent of an angle is given by the ratio of the cosine to the sine of that angle:
[tex]\[\cot^2\left(\frac{x}{2}\right) = \frac{\cos^2\left(\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right)}.\][/tex]
### Step 2: Use the double-angle identity for cosine
Using the double-angle identity for cosine, we can write:
[tex]\[\cos(x) = \cos(2 \theta) = 2 \cos^2(\theta) - 1,\][/tex]
where [tex]\(\theta = \frac{x}{2}\)[/tex]. Therefore:
[tex]\[\cos(x) = 2 \cos^2\left(\frac{x}{2}\right) - 1.\][/tex]
### Step 3: Express [tex]\(\cos^2\left(\frac{x}{2}\right)\)[/tex] and [tex]\(\sin^2\left(\frac{x}{2}\right)\)[/tex]
From the double-angle identity mentioned, isolate [tex]\(\cos^2\left(\frac{x}{2}\right)\)[/tex]:
[tex]\[2\cos^2\left(\frac{x}{2}\right) = \cos(x) + 1 \Rightarrow \cos^2\left(\frac{x}{2}\right) = \frac{1 + \cos(x)}{2}.\][/tex]
Similarly, using the Pythagorean identity [tex]\(\sin^2(\theta) + \cos^2(\theta) = 1\)[/tex]:
[tex]\[\sin^2\left(\frac{x}{2}\right) = 1 - \cos^2\left(\frac{x}{2}\right).\][/tex]
Now substitute [tex]\(\cos^2\left(\frac{x}{2}\right)\)[/tex] from above:
[tex]\[\sin^2\left(\frac{x}{2}\right) = 1 - \frac{1 + \cos(x)}{2} = \frac{2 - (1 + \cos(x))}{2} = \frac{1 - \cos(x)}{2}.\][/tex]
### Step 4: Substitute back into the cotangent identity
Now, substitute the expressions for [tex]\(\cos^2\left(\frac{x}{2}\right)\)[/tex] and [tex]\(\sin^2\left(\frac{x}{2}\right)\)[/tex] back into the cotangent identity:
[tex]\[\cot^2\left(\frac{x}{2}\right) = \frac{\cos^2\left(\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right)} = \frac{\frac{1 + \cos(x)}{2}}{\frac{1 - \cos(x)}{2}} = \frac{1 + \cos(x)}{1 - \cos(x)}.\][/tex]
### Step 5: Result
We see that:
[tex]\[\cot^2\left(\frac{x}{2}\right) = \frac{\cos x + 1}{\cos x - 1},\][/tex]
which completes our proof and verifies the given identity.
So, we have verified that:
[tex]\[\boxed{\cot^2 \left(\frac{x}{2}\right) = \frac{\cos x + 1}{\cos x - 1}}\][/tex]
