Giải thích các bước giải:

Ta có:

$A=\dfrac{\tan^2x-\sin^2x+\cot^2x-\cos^2x}{\sin^22x}$

$\to A=\dfrac{\dfrac{\sin^2x}{\cos^2x}-\sin^2x+\dfrac{\cos^2x}{\sin^2x}-\cos^2x}{\sin^22x}$

$\to A=\dfrac{\dfrac{\sin^2x}{\cos^2x}+\dfrac{\cos^2x}{\sin^2x}-(\sin^2x+\cos^2x)}{\sin^22x}$

$\to A=\dfrac{\dfrac{\sin^2x}{\cos^2x}+\dfrac{\cos^2x}{\sin^2x}-1}{\sin^22x}$

$\to A=\dfrac{\dfrac{\sin^2x}{\cos^2x}+\dfrac{\cos^2x}{\sin^2x}+2-3}{\sin^22x}$

$\to A=\dfrac{(\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x})^2-3}{\sin^22x}$

$\to A=\dfrac{(\dfrac{\sin^2x+\cos^2x}{\sin x\cos x})^2-3}{\sin^22x}$

$\to A=\dfrac{(\dfrac{1}{\sin x\cos x})^2-3}{\sin^22x}$

$\to A=\dfrac{(\dfrac{2}{2\sin x\cos x})^2-3}{\sin^22x}$

$\to A=\dfrac{(\dfrac{2}{\sin2x})^2-3}{\sin^22x}$

$\to A=\dfrac{\dfrac4{\sin^22x}-3}{\sin^22x}$

(xem lại đề)