You are on the right track.
writing $\tan\theta$ as$ \dfrac {\sin\theta}{\cos\theta}$ and $\cot\theta$ as $ \dfrac {\cos\theta}{\sin\theta} $, we get
$ \dfrac {\frac {\sin\theta}{\cos\theta} }{1-\frac {\cos\theta}{\sin\theta} }+\dfrac {\frac {\cos\theta}{\sin\theta} }{1-\frac {\sin\theta}{\cos\theta} }$
$= \dfrac {\sin^2\theta}{cos\theta\cdot(\sin\theta-\cos\theta)} + \dfrac {\cos^2\theta}{\sin\theta\cdot(\cos\theta-\sin\theta)}$ (how?)
$= \dfrac {\sin^2\theta}{\cos\theta\cdot(\sin\theta-\cos\theta)} - \dfrac {\cos^2\theta}{\sin\theta\cdot(\sin\theta-\cos\theta)}$
$=\dfrac{1}{(\sin\theta-\cos\theta)}\big(\dfrac {\sin^2\theta}{\cos\theta}-\dfrac {\cos^2\theta}{\sin\theta})$
$=\dfrac{1}{(\sin\theta-\cos\theta)}\big(\dfrac {\sin^3\theta-\cos^3\theta}{\sin\theta\cdot\cos\theta})$
$=\dfrac{\sin\theta-\cos\theta}{\sin\theta-\cos\theta}\dfrac{\big(\sin^2\theta+\sin\theta\cdot\cos\theta+\cos^2\theta)}{\sin\theta\cdot\cos\theta}$(how?)
$=1\cdot \dfrac{1+\sin\theta\cdot\cos\theta}{\sin\theta\cdot\cos\theta}$ (why?)
which is
$1+\sec\theta\cdot\csc\theta$
QED.
Best Answer
$(\csc\theta-\sec\theta)(\cot\theta-\tan\theta)=\displaystyle\big(\frac{1}{\sin\theta}-\frac{1}{\cos\theta}\big)\big(\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}\big)$
$\displaystyle=\big(\frac{\cos\theta-\sin\theta}{\sin\theta\cos\theta}\big)\big(\frac{\cos^{2}\theta-\sin^2\theta}{\cos\theta\sin\theta}\big)$
$\displaystyle=\frac{(\cos\theta-\sin\theta)^2(\cos\theta+\sin\theta)}{\sin^2\theta\cos^2\theta}$
$\displaystyle=\frac{(\cos^2\theta-2\sin\theta\cos\theta+\sin^2\theta)(\cos\theta+\sin\theta)}{\sin^2\theta\cos^2\theta}$
$\displaystyle=\frac{(1-2\sin\theta\cos\theta)(\cos\theta+\sin\theta\big)}{\sin^2\theta\cos^2\theta}$
$\displaystyle=\big(\frac{\cos\theta+\sin\theta}{\sin\theta\cos\theta}\big)\big(\frac{1-2\sin\theta\cos\theta}{\sin\theta\cos\theta}\big)$
$\displaystyle=\big(\frac{1}{\sin\theta}+\frac{1}{\cos\theta}\big)\big(\frac{1}{\sin\theta\cos\theta}-2\big)$
$\displaystyle=(\csc\theta+\sec\theta)(\csc\theta\sec\theta-2)$