$\begin{array}{l}a)\,\tan^2x - sin^2x\\ =\dfrac{\sin^2x}{\cos^2x} -\sin^2x\\ = \dfrac{\sin^2x-\sin^2x\cos^2x}{\cos^2x}\\ =\dfrac{\sin^2x(1 - \cos^2x)}{\cos^2x}\\ =\dfrac{\sin^2x.\sin^2x}{\cos^2x}\\ = \tan^2x.\sin^2x\\ b)\,\dfrac{1-4\sin^2x\cos^2x}{(\sin x - \cos x)^2}\\ =\dfrac{\sin^2x - 2\sin^2x\cos^2x + \cos^2x - 2\sin^2x\cos^2x}{(\sin x - \cos x)^2}\\ = \dfrac{\sin^2x(1 - 2\cos^2x) + \cos^2x(1 - 2\sin^2x)}{(\sin x - \cos x)^2}\\ = \dfrac{\sin^2x(\sin^2x - \cos^2x) + \cos^2x(\cos^2x - \sin^2x)}{(\sin x - \cos x)^2}\\ = \dfrac{(\sin^2x - \cos^2x)^2}{(\sin x - \cos x)^2}\\ = \dfrac{(\sin x - \cos x)^2(\sin x + \cos x)^2}{(\sin x - \cos x)^2}\\ = (\sin x + \cos x)^2\\ c)\,\dfrac{1 - 2\sin x\cos x}{\sin^2x - \cos^2x}\\ = \dfrac{\sin^2x - \sin x \cos x + \cos^2x - \sin x\cos x}{\sin^2x - \cos^2x}\\ = \dfrac{\sin x(\sin x - \cos x)-\cos x(\sin x - \cos x)}{(\sin x - \cos x)(\sin x + \cos x}\\ = \dfrac{(\sin x - \cos x)^2}{(\sin x - \cos x)(\sin x + \cos x)}\\ = \dfrac{\sin x + \cos x}{\sin x - \cos x}\\ d)\, \dfrac{1 - 2\cos^2x}{1 + 2\sin x\cos x}\\ = \dfrac{\sin^2x + \cos^2x - 2\cos^2x}{\sin^2x +2\sin x\cos x+\cos^2x}\\ = \dfrac{\sin^2x - \cos^2x}{(\sin x + \cos x)^2}\\ = \dfrac{(\sin x - \cos x)(\sin x + \cos x)}{(\sin x + \cos x)^2}\\ =\dfrac{\sin x - \cos x}{\sin x + \cos x}\end{array}$
