$\frac{1}{\tan A+\cot A}$ is equal to:

Let's simplify the given expression first:
$$\frac{1}{\tan A+\cot A}$$
Express $\tan A$ and $\cot A$ in terms of $\sin A$ and $\cos A$:
$$=\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}$$
$$=\frac{1}{\frac{{\sin }^{2}A+{\cos }^{2}A}{\sin A\cos A}}$$
Since ${\sin }^{2}A+{\cos }^{2}A=1$:
$$=\frac{1}{\frac{1}{\sin A\cos A}}=\sin A\cos A\text{— (1)}$$
Now, let's simplify Option B:
$$\text{Expression}=(\csc A-\sin A)(\sec A-\cos A)$$
$$=\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)$$
$$=\left(\frac{1-{\sin }^{2}A}{\sin A}\right)\left(\frac{1-{\cos }^{2}A}{\cos A}\right)$$
Using the identity $1-{\sin }^{2}A={\cos }^{2}A$ and $1-{\cos }^{2}A={\sin }^{2}A$:
$$=\left(\frac{{\cos }^{2}A}{\sin A}\right)\left(\frac{{\sin }^{2}A}{\cos A}\right)$$
$$=\cos A\sin A$$
This perfectly matches our simplified term from equation (1).