Đáp án:
$0$
Giải thích các bước giải:
$VT=\dfrac{2\cos2x-\sin4a}{2\cos2a+\sin4a}\\
=\dfrac{2\cos2x-2\sin2a\cos2x}{2\cos2a+2\sin2a\cos2a}\\
=\dfrac{2\cos2x(1-\sin2a)}{2\cos2a(1+\sin2a)}\\
=\dfrac{1-\sin2a}{1+\sin2a}\\
VP=\tan^2\left ( \dfrac{\pi}{4}-a \right )\\
=\dfrac{1}{\cos^2\left ( \dfrac{\pi}{4}-a \right )}-1\\
=\dfrac{2}{\cos\left ( 2\dfrac{\pi}{4}-2a \right )+1}-1\\
=\dfrac{2}{\cos\left ( \dfrac{\pi}{2}-2a \right )+1}-1\\
=\dfrac{2}{\sin2a+1}-1\\
=\dfrac{2-\sin2a-1}{\sin2a+1}\\
=\dfrac{1-\sin2a}{\sin2a+1}\\
\Rightarrow \dfrac{2\cos2x-\sin4a}{2\cos2a+\sin4a}-\tan^2\left ( \dfrac{\pi}{4}-a \right )\\
=\dfrac{1-\sin2a}{1+\sin2a}-\dfrac{1-\sin2a}{\sin2a+1}=0$
