Giải thích các bước giải:
$0 < x <\dfrac{\pi}{2}\\ \Rightarrow \sin x>0; \cos x>0\\ E=\dfrac{\sin x}{\sin x-\sqrt{\cot^2x-\cos^2x}}-\dfrac{\cos^2x}{2\sin^2x-1}\\ =\dfrac{\sin x}{\sin x-\sqrt{\dfrac{\cos^2x}{\sin^2x}-\cos^2x}}-\dfrac{\cos^2x}{2\sin^2x-(\sin^2x+\cos^2x)}\\ =\dfrac{\sin x}{\sin x-\sqrt{\dfrac{\cos^2x-\cos^2x\sin^2x}{\sin^2x}}}-\dfrac{\cos^2x}{\sin^2x-\cos^2x}\\ =\dfrac{\sin x}{\sin x-\sqrt{\dfrac{\cos^2x(1-\sin^2x)}{\sin^2x}}}-\dfrac{\cos^2x}{\sin^2x-\cos^2x}\\ =\dfrac{\sin x}{\sin x-\sqrt{\dfrac{\cos^4x}{\sin^2x}}}-\dfrac{\cos^2x}{\sin^2x-\cos^2x}\\ =\dfrac{\sin x}{\sin x-\dfrac{\cos^2x}{\sin x}}-\dfrac{\cos^2x}{\sin^2x-\cos^2x}\\ =\dfrac{1}{1-\dfrac{\cos^2x}{\sin^2x}}-\dfrac{\cos^2x}{\sin^2x-\cos^2x}\\ =\dfrac{1}{\dfrac{\sin^2x-\cos^2x}{\sin^2x}}-\dfrac{\cos^2x}{\sin^2x-\cos^2x}\\ =\dfrac{\sin^2x}{\sin^2x-\cos^2x}-\dfrac{\cos^2x}{\sin^2x-\cos^2x}\\ =\dfrac{\sin^2x-\cos^2x}{\sin^2x-\cos^2x}\\ =1$
Vậy $E$ không phụ thuộc vào biến số.
