Giải thích các bước giải:
$VT=\cot^2x+\tan^2x\\
=\dfrac{\cos^2x}{\sin^2x}+\dfrac{\sin^2x}{\cos^2x}\\
=\dfrac{\cos^4x+\sin^4x}{\sin^2x\cos^2x}\\
=\dfrac{(\cos^2x+\sin^2x)^2-2\sin^2x\cos^2x}{\dfrac{1}{4}\sin^22x}\\
=\dfrac{1-2.\dfrac{1}{4}\sin^22x}{\dfrac{1}{4}\sin^22x}\\
=\dfrac{4(1-\dfrac{1}{2}\sin^22x)}{\sin^22x}\\
=\dfrac{4-2\sin^22x}{\sin^22x}=VP\Rightarrow ĐPCM$