Giải thích các bước giải:
$VP=\tan\left ( 2x+\dfrac{\pi}{4} \right )=\dfrac{\tan2x+\tan\dfrac{\pi}{4}}{1-\tan2x\tan\dfrac{\pi}{4}}\\
=\dfrac{\tan2x+1}{1-\tan2x}\\
=\dfrac{\dfrac{\sin2x}{\cos2x}+1}{1-\dfrac{\sin2x}{\cos2x}}\\
=\dfrac{\dfrac{\sin2x}{\cos2x}+\dfrac{\cos2x}{\cos2x}}{\dfrac{\cos2x}{\cos2x}-\dfrac{\sin2x}{\cos2x}}\\
=\dfrac{\sin2x+\cos2x}{\cos2x-\sin2x}\\$
$VT=\dfrac{1+2\sin2x\cos2x}{2\cos\left ( 2x+\dfrac{\pi}{4} \right )\cos\left ( 2x-\dfrac{\pi}{4} \right )}\\
=\dfrac{\sin^22x+\cos^22x+2\sin2x\cos2x}{2.\dfrac{1}{2}\left [\cos\left ( 2x+\dfrac{\pi}{4}-2x+\dfrac{\pi}{4} \right )+\cos\left (2x+\dfrac{\pi}{4}+ 2x-\dfrac{\pi}{4} \right ) \right ]}\\
=\dfrac{(\sin2x+\cos2x)^2}{\cos\dfrac{\pi}{2}+\cos4x}\\
=\dfrac{(\sin2x+\cos2x)^2}{\cos4x}\\
=\dfrac{(\sin2x+\cos2x)^2}{\cos^22x-\sin^22x}\\
=\dfrac{(\sin2x+\cos2x)^2}{(\cos2x-\sin2x)(\cos2x+\sin2x)}\\
=\dfrac{\sin2x+\cos2x}{\cos2x-\sin2x}=VP(đpcm$
