Đáp án:
$\begin{array}{l}
\dfrac{1}{{\sin 2x}} + \dfrac{1}{{\sin 4x}} + \dfrac{1}{{\sin 8x}} + \cot 8x\\
= \dfrac{1}{{\sin 2x}} + \dfrac{1}{{\sin 4x}} + \dfrac{1}{{\sin 8x}} + \dfrac{{\cos 8x}}{{\sin 8x}}\\
= \dfrac{1}{{\sin 2x}} + \dfrac{1}{{\sin 4x}} + \dfrac{{\cos 8x + 1}}{{\sin 8x}}\\
= \dfrac{1}{{\sin 2x}} + \dfrac{1}{{\sin 4x}} + \dfrac{{2{{\cos }^2}4x}}{{2.\cos 4x.\sin 4x}}\\
= \dfrac{1}{{\sin 2x}} + \dfrac{1}{{\sin 4x}} + \dfrac{{\cos 4x}}{{\sin 4x}}\\
= \dfrac{1}{{\sin 2x}} + \dfrac{{\cos 4x + 1}}{{\sin 4x}}\\
= \dfrac{1}{{\sin 2x}} + \dfrac{{2{{\cos }^2}2x}}{{2.\cos 2x.\sin 2x}}\\
= \dfrac{1}{{\sin 2x}} + \dfrac{{\cos 2x}}{{\sin 2x}}\\
= \dfrac{{2{{\cos }^2}x}}{{2.\cos x.\sin x}}\\
= \dfrac{{\cos x}}{{\sin x}}\\
= \cot x.
\end{array}$
