Solve identity : $\frac{1+\sin x\cos x}{\cos^3 x - \sin^3 x}+\frac{1}{\sin x+\cos x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}=\frac{1}{\tan^2x-1}$ $$\frac{1+\sin x\cos x}{\cos^3 x - \sin^3 x}+\frac{1}{\sin x+\cos x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}=\frac{1}{\tan^2x-1}$$
I am doing this identity for about an hour and I can't get to the result. I use a common denominator but that just makes a mess that takes forever to do and gets me nowhere. Surely there is an easier way that I am not seeing? Some hint or help please?

Use  $a^3-b^3=(a-b)(a^2+ab+b^2)$
Therefore we have, 
$$\cos^3x-\sin^3x=(\cos x-\sin x)(1+\sin x \cos x)$$
Your expression in LHS
$$\frac{1+\sin x\cos x}{\cos^3 x - \sin^3 x}+\frac{1}{\sin x+\cos x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x} $$
Simplifies to 
\begin{align}
\text{LHS} 
&=\frac{1}{\cos x - \sin x}+\frac{1}{\sin x+\cos x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}\\
&= \frac{2\cos x}{\cos^2 x - \sin^2 x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}\\
&= \frac{2 \cos x+\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}\\
&= \frac{\sin^2 x-1}{\cos^2x-\sin^2x}\\
&= \frac{cos^2 x}{\sin ^2x-\cos^2x}\\
&= \frac{1}{\tan^2x-1}= \text{RHS}\\
\end{align}
