How do you simplify $ \sin x\cot x $ ?

Hint: We first try to find the respective ratios of the multiplication $ \sin x\cot x $ . We find their respective values according to the sides of a right-angle triangle. We use the relations $ \sin x=\dfrac{height}{hypotenuse} $ and $ \cot x=\dfrac{base}{height} $ to multiply them. Then from the final ration we find the solution.

Complete step by step answer:
The given trigonometric expression is the multiplication of two ratios $ \sin x $ and $ \cot x $ .
We have their respective values according to the sides of a right-angle triangle. We use those relations to find the value of $ \sin x\cot x $ .
According to a right-angle triangle, the value of $ \sin x $ will be considered as the ratio of the length of the height to the hypotenuse with respect to a certain angle.
So, $ \sin x=\dfrac{height}{hypotenuse} $ .
And according to the same right-angle triangle, the value of $ \cot x $ will be considered as the ratio of the length of the base to the height with respect to the same angle.
So, $ \cot x=\dfrac{base}{height} $ .
The multiplied form of the term $ \sin x\cot x $ gives $ \sin x\cot x=\dfrac{height}{hypotenuse}\times \dfrac{base}{height}=\dfrac{base}{hypotenuse} $ .
The ratio of $ \dfrac{base}{hypotenuse} $ is defined as $ \dfrac{base}{hypotenuse}=\cos x $.
Therefore, $ \sin x\cot x=\cos x $ .

Note:
 We can also use the direct ratio relation to find the solution. We know that $ \cot x $ can be broken into ratio of two other trigonometric expression which are $ \cos x $ and $ \sin x $ . We know that $ \cot x=\dfrac{\cos x}{\sin x} $ . Now we multiply $ \sin x $ on both sides of the equation and get
\[\begin{align}
  & \cot x\times \sin x=\dfrac{\cos x}{\sin x}\times \sin x \\
 & \Rightarrow \sin x\cot x=\cos x \\
\end{align}\].
Thus, verified the relation $ \sin x\cot x=\cos x $ .