Trigonometric identity, simplifying an expression to $(1-\sin^2 a\cos^2a)/(2+\sin^2a\cos^2a)$ Question:
$$\left(\frac{1}{\sec^2A-\cos^2A}+\frac{1}{\csc^2A -\sin^2A}\right)\sin^2A\cos^2A=\frac{1-\sin^2A \cos^2A}{2+\sin^2A\ \cos^2A}$$
Prove L.H.S. = R.H.S.
My Efforts:
$$=\left(\frac{1}{\frac{1}{\cos^2 A}-\cos^2A}+\frac{1}{\frac{1}{\sin^2A} -\sin^2A}\right)\sin^2A\cos^2A$$
$$=\left(\frac{\cos^2A}{1-\cos^4A}+\frac{\sin^2A}{1-\sin^4A }\right)\sin^2A\cos^2A$$
I thought i will reach to R.H.S. by what i was doing, but now i am stuck I don't how to proceed further or is there a way to reach R.H.S. with what I am doing. 
 A: $$\left(\frac{\cos^2A}{1-\cos^4A}+\frac{\sin^2A}{1-\sin^4A }\right)\sin^2A\cos^2A$$
$$=\left(\frac{\cos^2A}{(1+\cos^2A)(1-\cos^2A)}+\frac{\sin^2A}{(1+\sin^2A)(1-\sin^2A) }\right)\sin^2A\cos^2A$$
$$=\left(\frac{\cos^2A}{(1+\cos^2A)\sin^2A}+\frac{\sin^2A}{(1+\sin^2A)\cos^2A}\right)\sin^2A\cos^2A$$
$$=\frac{\cos^4A}{1+\cos^2A}+\frac{\sin^4A}{1+\sin^2A}=\frac{\color{red}{\cos^4A}+\cos^4A\sin^2A\color{red}{+\sin^4A}+\sin^4A\cos^2A}{(1+\cos^2A)(1+\sin^2A)}$$
Here, the numerator will be
$$\color{red}{(\cos^2A+\sin^2A)^2-2\cos^2A\sin^2A}+\cos^2A\sin^2A(\cos^2A+\sin^2A)=1-\cos^2A\sin^2A.$$
On the other hand, the denominator will be
$$1+\sin^2A+\cos^2A+\cos^2A\sin^2A=2+\cos^2A\sin^2A.$$
(Maybe I wrote too much...)
A: Next step: $1-\cos^4A=(1+\cos^2A)(1-\cos^2A)=(1+\cos^2A)\sin^2A$
A: $$\sin^2A\cos^2A\left(\frac1{\sec^2A-\cos^2A}\right)=\sin^2A\cos^2A\left(\frac{\cos^2A}{(1+\cos^2A)\sin^2A}\right)$$
$$=\frac{\cos^4A}{1+\cos^2A} =\frac{\cos^4A-1+1}{1+\cos^2A} =\cos^2A-1+\frac1{1+\cos^2A}$$
Similarly, $$\sin^2A\cos^2A\left(\frac1{\csc^2A-\sin^2A}\right)=\sin^2A-1+\frac1{1+\sin^2A}$$
Adding we get $$\cos^2A+\sin^2A-1-1+\frac1{1+\cos^2A}+\frac1{1+\sin^2A}$$
$$=-1+\frac{(1+\cos^2A)+(1+\sin^2A)}{1+\cos^2A+\sin^2A+\cos^2A\sin^2A}$$
$$=-1+\frac3{2+\cos^2A\sin^2A}=\cdots$$
