Be careful!

If $A = \frac{\pi}{4}$ , then we have:

$\begin{aligned} LHS & = \left( \frac{1}{\sec^2 \frac{\pi}{4} - \cos^2 \frac{\pi}{4}} + \frac{1}{\csc^2 \frac{\pi}{4} - \sin^2 \frac{\pi}{4}} \right) \cos^2 \frac{\pi}{4} \sin^2 \frac{\pi}{4} \\ & = \left( \frac{1}{2 - \frac{1}{2}} + \frac{1}{2 - \frac{1}{2}} \right) \frac{1}{2} \times \frac{1}{2} \\ & = \frac{1}{3} \\ RHS & = \frac{1-2 \cos^2 \frac{\pi}{4} \sin^2 \frac{\pi}{4}}{2+\cos^2 \frac{\pi}{4} \sin^2 \frac{\pi}{4}} \\ & = \frac{1-\frac{2}{4}}{2+\frac{1}{4}} \\ & = \frac{2}{9} \end{aligned}$

$\Rightarrow LHS \le RHS \quad \Rightarrow \text{The identity is } \boxed{false}$ .

Simplifying the LHS in terms of 'sin A' & 'cos A' and then solving further shows that the coefficient of the trigonometrical term in the numerator of the RHS is '-1', not '-2'. Hence disproven.