How do you verify $\cos x\cot x+\sin x=\csc x$ ?

Hint: For solving this problem, we first simplify the LHS of the equation and finally arrange it in such a way to make it look like the RHS of the equation, keeping all the steps correct. First, we write $\cot x$ as $\dfrac{\cos x}{\sin x}$ and then multiply it with $\cos x$ . Then, adding $\sin x$ to it makes the numerator ${{\sin }^{2}}x+{{\cos }^{2}}x$ and the denominator as $\sin x$ . Finally, we use the property of trigonometry ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ . This makes the numerator as $1$ with denominator as $\sin x$ . This evaluates to $\csc x$ .

Complete step-by-step answer:
The given equation is
$\cos x\cot x+\sin x=\csc x....equation1$
We verify the equation by simplifying the LHS of the equation and manipulating it to become equal to the RHS. The LHS is
$\cos x\cot x+\sin x$
We express $\cot x$ as $\dfrac{\cos x}{\sin x}$ . The LHS thus becomes,
$\Rightarrow \cos x\times \left( \dfrac{\cos x}{\sin x} \right)+\sin x$
Multiplying $\cos x$ with $\dfrac{\cos x}{\sin x}$ , we get
$\Rightarrow \left( \dfrac{{{\cos }^{2}}x}{\sin x} \right)+\sin x$
Adding the two terms, we get,
$\Rightarrow \left( \dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{\sin x} \right)$
We know that there is a common property in trigonometry which is ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ . Using it in the LHS, we get,
$\Rightarrow \dfrac{1}{\sin x}$
$\csc x$ is nothing but the reciprocal of $\sin x$ . Therefore, the expression becomes
$\Rightarrow \csc x$
This is nothing but the RHS of the $equation1$ . Thus, $LHS=RHS$ and hence, the given equation is verified.

Note: We need to remember the exact properties and should not interchange them like, $\cot x$ is $\dfrac{\cos x}{\sin x}$ and not the other way round. Also, we must multiply and add or subtract the terms correctly. In some cases RHS is complicated but LHS is a simplified expression. In that case, we simplify the RHS and manipulate it to look equal to the LHS.