a,
$A=\dfrac{ \sin^2x+2\sin x\cos x+\cos^2x-1}{ \cot x-\sin x\cos x}$
$=\dfrac{1+2\sin x\cos x-1}{ \dfrac{\cos x}{\sin x}-\sin x\cos x}$
$=\dfrac{2\sin x\cos x}{ \dfrac{\cos x-\sin^2x\cos x}{\sin x} }$
$=\dfrac{2\sin^2x\cos x}{\cos x(1-\sin^2x)}$
$=\dfrac{2\sin^2x\cos x}{\cos^3x}$
$=2\tan^2x$
b,
$B=\dfrac{\sin^2x+\cos^2x-1}{\cot^2x}$
$=\dfrac{1-1}{\cot^2x}$
$=0$
c,
$C=\dfrac{\sin^2x-\tan^2x}{\cos^2x-\cot^2x}$
$=\dfrac{ \sin^2x-\dfrac{\sin^2x}{\cos^2x} }{\cos^2x-\dfrac{\cos^2x}{\sin^2x} }$
$=\dfrac{\sin^2x\cos^2x-\sin^2x}{\cos^2x}: \dfrac{ \cos^2x\sin^2x-\cos^2x}{\sin^2x}$
$=\dfrac{\sin^2x(\cos^2x-1)}{\cos^2x}.\dfrac{\sin^2x}{\cos^2x(\sin^2x-1)}$
$=\dfrac{\sin^2x.(-\sin^2x).\sin^2x}{\cos^2x.\cos^2x.(-\cos^2x)}$
$=\tan^6x$
d,
$D=\dfrac{\cot^2x-\cos^2x}{\cos^2x}+1-\cos^2x$
$=\dfrac{\cot^2x}{\cos^2x}-1+1-\cos^2x$
$=\dfrac{\cos^2x}{\sin^2x\cos^2x}-\cos^2x$
$=\dfrac{1}{\sin^2x}-\cos^2x$
$=1+\cot^2x-\cos^2x$
$=\cot^2x+\sin^2x$
