4 (a) Show that $\frac{\mathrm{d}}{\mathrm{d} x}(\cot x)=-\operatorname{cosec}^2 x$.
[2]
(b) Show that $\sin 2 x \tan x=2 \sin ^2 x$.
[2]
(c) Hence, find the exact value of $\int_{\frac{\pi}{18}}^{\frac{\pi}{9}} \operatorname{cosec} 6 x \cot 3 x \mathrm{~d} x$.
[4]
(c) $\frac{\sqrt{3}}{9}$
(a)
$$
\begin{array}{l}
\frac{d}{d x} \cot x \\
=\frac{d}{d x} \frac{\cos x}{\sin x} \\
=\frac{-\sin x \cdot \sin x-\cos x \cdot \cos x}{\sin ^2 x} \\
=\frac{-1}{\sin ^2 x} \\
=-\operatorname{cosec}^2 x
\end{array}
$$
(b)
$$
\begin{array}{l}
\sin 2 x \tan x \\
=2 \sin x \cos x \cdot \frac{\sin x}{\cos x} \\
=2 \sin ^2 x
\end{array}
$$
(c)
$$
\begin{array}{l}
\int \operatorname{cosec} 6 x \cot 3 x d x \\
=\int \frac{1}{\sin 6 x \tan 3 x} d x \\
=\int \frac{1}{2 \sin ^2 3 x} d x \\
=\frac{1}{2} \int \operatorname{cosec}^2 3 x d x \\
=-\frac{1}{6} \cot 3 x+c \\
\int_{\pi / 18}^{\pi / 9} \operatorname{cosec} 6 x \cot 3 x d x \\
=-\frac{1}{6}[\cot 3 x]_{\pi / 18}^{\pi / 9} \\
=-\frac{1}{6}\left(\cot \frac{\pi}{3}-\cot \frac{\pi}{6}\right)\\
=\frac{\sqrt{3}}{9}
\end{array}
$$