Step 1: Take the LHS and write $\operatorname{cosec} A=\dfrac{1}{\sin A}$, $\sec A=\dfrac{1}{\cos A}$.
$$\text{LHS}=\left(\sin A-\frac{1}{\sin A}\right)\left(\cos A-\frac{1}{\cos A}\right)$$
Step 2: Combine each bracket into a single fraction.
$$=\left(\frac{\sin^{2}A-1}{\sin A}\right)\left(\frac{\cos^{2}A-1}{\cos A}\right)$$
Step 3: Use $\sin^{2}A-1=-\cos^{2}A$ and $\cos^{2}A-1=-\sin^{2}A$.
$$=\left(\frac{-\cos^{2}A}{\sin A}\right)\left(\frac{-\sin^{2}A}{\cos A}\right)=\frac{\cos^{2}A\,\sin^{2}A}{\sin A\,\cos A}$$
Step 4: Cancel to get
$$\text{LHS}=\sin A\cos A$$
Step 5: Now simplify the RHS.
$$\tan A+\cot A=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}=\frac{\sin^{2}A+\cos^{2}A}{\sin A\cos A}=\frac{1}{\sin A\cos A}$$
Step 6: Therefore
$$\text{RHS}=\frac{1}{\tan A+\cot A}=\sin A\cos A$$
Step 7: LHS $=$ RHS $=\sin A\cos A$. Hence proved.