Step 1: Take the LHS and expand both squares.
$$\text{LHS} = \sin^2 A + \sec^2 A + 2\sin A\sec A + \cos^2 A + \text{cosec}^2 A + 2\cos A\,\text{cosec }A$$
Step 2: Group $\sin^2 A + \cos^2 A = 1$.
$$= 1 + (\sec^2 A + \text{cosec}^2 A) + 2\left(\sin A \cdot \frac{1}{\cos A} + \cos A \cdot \frac{1}{\sin A}\right)$$
Step 3: Simplify the bracket.
$$\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2 A + \cos^2 A}{\sin A\cos A} = \frac{1}{\sin A\cos A}$$
Step 4: Also, $$\sec^2 A + \text{cosec}^2 A = \frac{1}{\cos^2 A} + \frac{1}{\sin^2 A} = \frac{\sin^2 A + \cos^2 A}{\sin^2 A\cos^2 A} = \frac{1}{\sin^2 A\cos^2 A}$$
Step 5: Therefore
$$\text{LHS} = 1 + \frac{1}{\sin^2 A\cos^2 A} + \frac{2}{\sin A\cos A}$$
Step 6: Note that $\sec A\ \text{cosec }A = \dfrac{1}{\cos A}\cdot\dfrac{1}{\sin A} = \dfrac{1}{\sin A\cos A}$.
Let $t = \sec A\ \text{cosec }A$. Then
$$\text{LHS} = 1 + t^2 + 2t = (1 + t)^2$$
Step 7: $$\text{LHS} = (1 + \sec A\ \text{cosec }A)^2 = \text{RHS}$$
Hence proved.