The expression shown in the image is:

$\sin(\cot^{-1} x)$

This represents the sine of the inverse cotangent (or arccotangent) of $x$. Let's simplify it.

Step-by-Step Solution:

1. Let $\theta = \cot^{-1} x$. This means: $\cot(\theta) = x$ From the definition of cotangent: $\cot(\theta) = \frac{1}{\tan(\theta)} = x \quad \text{or} \quad \tan(\theta) = \frac{1}{x}.$

2. We need to find $\sin(\theta)$. We can use a right triangle to represent this. If $\tan(\theta) = \frac{1}{x}$, then in the corresponding right triangle:

   • The opposite side is 1.
   • The adjacent side is $x$.
   • The hypotenuse (by the Pythagorean theorem) is: $\text{Hypotenuse} = \sqrt{1^2 + x^2} = \sqrt{x^2 + 1}.$

3. Now, using the definition of sine: $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2 + 1}}.$

Thus, the simplified result of $\sin(\cot^{-1} x)$ is:

$\sin(\cot^{-1} x) = \frac{1}{\sqrt{x^2 + 1}}.$

Would you like more details on the derivation or have any other questions?

Here are 5 related questions to consider:

1. How do you derive the expression for $\cos(\cot^{-1} x)$?
2. What is the relationship between the inverse cotangent and inverse tangent functions?
3. How do you compute the derivative of $\sin(\cot^{-1} x)$?
4. How do you express $\cot^{-1} x$ in terms of $\arctan(x)$?
5. What are the domain and range of $\sin(\cot^{-1} x)$?

Tip: For solving inverse trigonometric functions, using right triangles or their unit circle equivalents often simplifies the work.