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
\begin{align} \sin A \frac{(\sin A+\cos A)}{\cos A}&+\cos A\frac{(\sin A+\cos A)}{\sin A} \\[6px]&= (\sin A+\cos A)\frac{\sin A}{\cos A}+(\sin A+\cos A)\frac{\cos A}{\sin A} \\[6px]&= (\cos A+\sin A)\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right) \\[6px]&= (\cos A+\sin A)\frac{(\sin^2 A+\cos^2 A)}{(\cos A\sin A)} \\[6px]&= \frac{\cos A}{\cos A\sin A}+\frac{\sin A}{\cos A\sin A} \\[6px]&= \frac1{\sin A}+\frac1{\cos A} \\[6px]&= \csc A+\sec A \end{align}