## Giàu giúp mình với..

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Giàu giúp mình với..

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9 months 2021-05-10T04:48:54+00:00 1 Answers 4 views 0

$\begin{array}{l} a)\cot x – \tan x – 2\tan 2x\\ = \dfrac{{\cos x}}{{\sin x}} – \dfrac{{\sin x}}{{\cos x}} – \dfrac{{2\sin 2x}}{{\cos 2x}}\\ = \dfrac{{{{\cos }^2}x – {{\sin }^2}x}}{{\sin x\cos x}} – \dfrac{{2\sin 2x}}{{\cos 2x}}\\ = \dfrac{{2\cos 2x}}{{\sin 2x}} – \dfrac{{2\sin 2x}}{{\cos 2x}}\\ = \dfrac{{2\left( {{{\cos }^2}2x – {{\sin }^2}2x} \right)}}{{\sin 2x\cos 2x}}\\ = \dfrac{{2\cos 4x}}{{\sin 2x\cos 2x}}\\ = \dfrac{{4\cos 4x}}{{\sin 4x}}\\ = 4\cot 4x\\ b)\dfrac{{1 – 2{{\sin }^2}2x}}{{1 – \sin 4x}}\\ = \dfrac{{{{\cos }^2}2x – {{\sin }^2}2x}}{{{{\cos }^2}2x – 2\sin 2x\cos 2x + {{\sin }^2}2x}}\\ = \dfrac{{\left( {\cos 2x – \sin 2x} \right)\left( {\cos 2x + \sin 2x} \right)}}{{{{\left( {\cos 2x – \sin 2x} \right)}^2}}}\\ = \dfrac{{\cos 2x + \sin 2x}}{{\cos 2x – \sin 2x}}\\ = \dfrac{{1 + \dfrac{{\sin 2x}}{{\cos 2x}}}}{{1 – \dfrac{{\sin 2x}}{{\cos 2x}}}}\\ = \dfrac{{1 + \tan 2x}}{{1 – \tan 2x}} \end{array}$