$\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}$ without L'Hôpitals Rule. The limit,
$\displaystyle{\lim \limits_{x \to \frac\pi4}\frac{(\csc^2x-2\tan^2x)}{(\cot x-1)}}=6$
is easily found using L'Hôpitals Rule. However, the exercise is to evaluate it without using this method. I've tried multiple substitutions, mainly pythagoric and reciprocal identities, but without success, the indeterminate form $\frac{0}{0}$ keeps appearing. I’ve also tried splitting the limit and doing the substitutions. I'm looking to simplify the expression in order to evaluate the limit directly.
Thanks in advance.

I use (at the end)
$\begin{array}\\
\cos x-\sin x
&=\sqrt{2}((1/\sqrt{2})\cos x-(1/\sqrt{2})\sin x)\\
&=\sqrt{2}(\sin(\pi/4)\cos x-\cos(\pi/4)\sin x)\\
&=\sqrt{2}\sin(\pi/4-x)\\
\end{array}
$
$\begin{array}\\
v
&=\lim \limits_{x \to \frac\pi4}\dfrac{\csc^2x-2\tan^2x}{\cot x-1}\\
&=\lim \limits_{x \to \frac\pi4}\dfrac{\dfrac1{\sin^2x}-2\dfrac{\sin^2x}{\cos^2x}}{\dfrac{\cos x}{\sin x}-1}\\
&=\lim \limits_{x \to \frac\pi4}\dfrac{\cos^2x-2\sin^4x}{\sin x\cos^3 x-\sin^2x\cos^2x}\\
&=\lim \limits_{x \to \frac\pi4}\dfrac{1-\sin^2x-2\sin^4x}{\sin x\cos x(1-\sin^2x)-\sin^2x(1-\sin^2x)}\\
&=\lim \limits_{x \to \frac\pi4}\dfrac{(1-2\sin^2x)(1+\sin^2x)}{\sin x(1-\sin^2x)(\cos x-\sin x)}\\
&=\lim \limits_{x \to \frac\pi4}\dfrac{(1-2\sin^2x)(3/2)}{(\sqrt{2}/2)(\cos x-\sin x)}\\
&=\dfrac{3}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{1-2\sin^2x}{(\cos x-\sin x)}\\
&=\dfrac{3}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{(1-\sqrt{2}\sin x)(1+\sqrt{2}\sin x)}{\cos x-\sin x}\\
&=\dfrac{3}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{(1-\sqrt{2}\sin x)(1+\sqrt{2}(\sqrt{2}/2))}{\cos x-\sin x}\\
&=\dfrac{6}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{(1-\sqrt{2}\sin x)}{\cos x-\sin x}\\
&=\dfrac{6}{\sqrt{2}}\lim \limits_{x \to \frac\pi4}\dfrac{1-\sqrt{2}\sin x}{\sqrt{2}\sin(\pi/4-x)}\\
&=3\lim \limits_{x \to \frac\pi4}\dfrac{1-\sqrt{2}\sin x}{\sin(\pi/4-x)}\\
&=3\lim \limits_{y \to 0}\dfrac{1-\sqrt{2}\sin (y+\pi/4)}{\sin(-y)}\\
&=3\lim \limits_{y \to 0}\dfrac{1-\sqrt{2}(\sin y\cos(\pi/4)+\cos(y)\sin(\pi/4))}{\sin(-y)}\\
&=3\lim \limits_{y \to 0}\dfrac{1-2(\sin y+\cos(y))}{\sin(-y)}\\
&=3\lim \limits_{y \to 0}\dfrac{-2\sin y}{-\sin(y)}\\
&=6\\
\end{array}
$
