Đáp án:
$12$
Giải thích các bước giải:
$K=\tan x+\cot x+\cot2x+\tan2x\\
=\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}+\dfrac{\cos2x}{\sin2x}+\dfrac{\sin2x}{\cos2x}\\
=\dfrac{\sin^2x}{\sin x\cos x}+\dfrac{\cos^2x}{\sin x\cos x}+\dfrac{\cos^22x}{\sin2x\cos2x}+\dfrac{\sin^22x}{\sin2x\cos2x}\\
=\dfrac{\sin^2x+\cos^2x}{\sin x\cos x}+\dfrac{\sin^22x+\cos^22x}{\sin2x\cos2x}\\
=\dfrac{1}{\dfrac{1}{2}\sin2x}+\dfrac{1}{\dfrac{1}{2}\sin4x}\\
=\dfrac{2}{\sin2x}+\dfrac{2}{\sin4x}\\
=\dfrac{2\sin4x+2\sin2x}{\sin2x\sin4x}\\
=\dfrac{2(\sin4x+\sin2x)}{\sin2x\sin4x}\\
=\dfrac{2.2\sin\dfrac{4x+2x}{2}\cos\dfrac{4x-2x}{2}}{\sin2x\sin4x}\\
=\dfrac{4\sin3x\cos x}{\sin2x\sin4x}\\
=\dfrac{4\sin3x\cos x}{2\sin x\cos x\sin4x}\\
=\dfrac{2\sin3x}{\sin x\sin4x}\\
\Rightarrow p=3,n=1,m=4\\
\Rightarrow m.n.p=4.1.3=12$
