To verify the given trigonometric identity:
[tex]\[ \csc^2(x) - 2 \csc(x) \cot(x) + \cot^2(x) = \tan^2\left(\frac{x}{2}\right) \][/tex]
we will simplify the left-hand side and see if it is equal to the right-hand side.
### Step-by-Step Solution:
1. Start with the left-hand side (LHS) of the equation:
[tex]\[ \csc^2(x) - 2 \csc(x) \cot(x) + \cot^2(x) \][/tex]
2. Express the trigonometric functions [tex]\(\csc(x)\)[/tex] and [tex]\(\cot(x)\)[/tex] in terms of [tex]\(\sin(x)\)[/tex] and [tex]\(\cos(x)\)[/tex]:
[tex]\[ \csc(x) = \frac{1}{\sin(x)} \][/tex]
[tex]\[ \cot(x) = \frac{\cos(x)}{\sin(x)} \][/tex]
3. Substitute these expressions into the LHS:
[tex]\[ \left(\frac{1}{\sin(x)}\right)^2 - 2 \left(\frac{1}{\sin(x)}\right) \left(\frac{\cos(x)}{\sin(x)}\right) + \left(\frac{\cos(x)}{\sin(x)}\right)^2 \][/tex]
4. Simplify each term:
[tex]\[ \frac{1}{\sin^2(x)} - 2 \frac{\cos(x)}{\sin^2(x)} + \frac{\cos^2(x)}{\sin^2(x)} \][/tex]
5. Combine the terms under a common denominator:
[tex]\[ \frac{1 - 2 \cos(x) + \cos^2(x)}{\sin^2(x)} \][/tex]
6. Factor the numerator:
Notice that [tex]\( 1 - 2 \cos(x) + \cos^2(x) \)[/tex] is a perfect square:
[tex]\[ 1 - 2 \cos(x) + \cos^2(x) = (\cos(x) - 1)^2 \][/tex]
7. Rewrite the LHS:
[tex]\[ \frac{(\cos(x) - 1)^2}{\sin^2(x)} \][/tex]
8. At this point, the LHS simplified form is:
[tex]\[ \frac{(\cos(x) - 1)^2}{\sin^2(x)} \][/tex]
Next, we examine the right-hand side (RHS):
9. The RHS of the equation is:
[tex]\[ \tan^2\left(\frac{x}{2}\right) \][/tex]
### Conclusion:
After simplifying the left-hand side, the equation becomes:
[tex]\[ \frac{(\cos(x) - 1)^2}{\sin^2(x)} \][/tex]
According to the identity, [tex]\(\tan^2\left(\frac{x}{2}\right) = \frac{(\cos(x) - 1)^2}{\sin(x)^2}\)[/tex], thus:
[tex]\[ \tan^2 \left( \frac{x}{2} \right) = \frac{(\cos(x) - 1)^2}{\sin^2(x)} \][/tex]
### Final Verification:
Therefore, the given trigonometric identity holds true:
[tex]\[ \csc^2(x) - 2 \csc(x) \cot(x) + \cot^2(x) = \tan^2\left( \frac{x}{2} \right) \][/tex]
We have verified that the left-hand side simplifies to the same form, proving that the identity is indeed correct.
