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.
Here is a simpler solution to this problem:
$$\left(\sec(\theta)+\tan(\theta) \right)\left(\sec(\theta)-\tan(\theta) \right)=\sec^2(\theta)-\tan^2(\theta)=1$$
Since $\sec(\theta)+\tan(\theta)=5$ you get $\sec(\theta)-\tan(\theta)=\frac{1}{5}$.
Adding and subtracting these two relations you get
$$2\sec(\theta)=5+\frac15=\frac{26}{5} \,;\, 2\tan(\theta)=5-\frac15=\frac{24}{5}$$
Thus $\tan(\theta)=\frac{24}{10}$ and
$$\sin(\theta)=\frac{\tan(\theta)}{\sec(\theta)}=\frac{24}{26} \,.$$
Best Answer
Write $1$ in the numerator as : $$\sec^2(\theta) - \tan^2(\theta)$$
$$\frac{(\sec\theta -\tan\theta)^2+\sec^2\theta - \tan^2\theta}{\sec\theta \csc\theta -\tan\theta \csc \theta} $$
$$\frac{(\sec\theta -\tan\theta)^2+(\sec\theta - \tan\theta)(\sec\theta + \tan\theta)}{\sec\theta \csc\theta -\tan\theta \csc \theta} $$
$$\frac{(\sec\theta - \tan\theta)(\sec\theta - \tan\theta + \sec\theta + \tan\theta)}{\sec\theta \csc\theta -\tan\theta \csc \theta} $$
$$\frac{(\sec\theta - \tan\theta)(2 \sec\theta)}{\csc\theta(\sec\theta - \tan\theta)}$$
$$2 \tan\theta$$
Hence the simplified result is: $$2 \tan\theta$$
Hope the answer is clear !