Step 1: Take the LHS and rationalise by multiplying numerator and denominator inside the root by $(\operatorname{cosec} A-1)$.
$$\text{LHS}=\sqrt{\frac{(\operatorname{cosec} A-1)}{(\operatorname{cosec} A+1)}\times\frac{(\operatorname{cosec} A-1)}{(\operatorname{cosec} A-1)}}$$
Step 2: This gives
$$=\sqrt{\frac{(\operatorname{cosec} A-1)^{2}}{\operatorname{cosec}^{2}A-1}}$$
Step 3: Use the identity $\operatorname{cosec}^{2}A-1=\cot^{2}A$.
$$=\sqrt{\frac{(\operatorname{cosec} A-1)^{2}}{\cot^{2}A}}=\frac{\operatorname{cosec} A-1}{\cot A}$$
Step 4: Write in terms of $\sin A$ and $\cos A$.
$$=\frac{\dfrac{1}{\sin A}-1}{\dfrac{\cos A}{\sin A}}=\frac{\dfrac{1-\sin A}{\sin A}}{\dfrac{\cos A}{\sin A}}$$
Step 5: Simplify.
$$=\frac{1-\sin A}{\cos A}=\frac{1}{\cos A}-\frac{\sin A}{\cos A}$$
Step 6: Therefore
$$=\sec A-\tan A=\text{RHS}$$
Hence proved.