Certainly! Let's walk through the verification of the trigonometric identity:
[tex]\[\cot ^2\left(\frac{x}{2}\right)=\frac{\cos x+1}{\cos x-1}\][/tex]
### Step-by-Step Solution:
1. Given:
[tex]\[
\cot ^2\left(\frac{x}{2}\right)
\][/tex]
This is our starting expression.
2. Transform Cotangent to Tangent Form:
[tex]\[
\cot\left(\frac{x}{2}\right) = \frac{1}{\tan\left(\frac{x}{2}\right)}
\][/tex]
Hence,
[tex]\[
\cot ^2\left(\frac{x}{2}\right) = \left(\frac{1}{\tan\left(\frac{x}{2}\right)}\right)^2
\][/tex]
3. Expressing [tex]\(\tan\left(\frac{x}{2}\right)\)[/tex]:
By using the half-angle formula for tangent,
[tex]\[
\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1 - \cos x}{1 + \cos x}}
\][/tex]
So,
[tex]\[
\cot ^2\left(\frac{x}{2}\right) = \left(\frac{1}{\sqrt{\frac{1 - \cos x}{1 + \cos x}}}\right)^2
\][/tex]
4. Simplify the Square Root Term:
[tex]\[
\cot ^2\left(\frac{x}{2}\right) = \frac{1}{\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}\right)^2}
\][/tex]
Since [tex]\(\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}\right)^2\)[/tex] simplifies to [tex]\(\frac{1 - \cos x}{1 + \cos x}\)[/tex],
[tex]\[
\cot ^2\left(\frac{x}{2}\right) = \frac{1}{\frac{1 - \cos x}{1 + \cos x}}
\][/tex]
5. Utilize the Multiplicative Inverse (Reciprocal):
By taking the reciprocal,
[tex]\[
\cot ^2\left(\frac{x}{2}\right) = \frac{1 + \cos x}{1 - \cos x}
\][/tex]
Thus, we have successfully shown that:
[tex]\[
\cot ^2\left(\frac{x}{2}\right) = \frac{\cos x + 1}{1 - \cos x}
\][/tex]
This completes the proof of the identity.
