We have to work with left-hand side and make it same as right-hand side.
Let's do it:
[tex]\frac{\sin \theta}{1-\cos \theta}[/tex][tex]\frac{\sin\theta}{1-\cos\theta}\times\frac{1+\cos\theta}{1+\cos\theta}=\frac{\sin \theta(1+\cos \theta)}{(1-\cos \theta)(1+\cos \theta)_{}}[/tex]Now, simplifying further:
[tex]\begin{gathered} \frac{\sin\theta(1+\cos\theta)}{(1-\cos\theta)(1+\cos\theta)_{}} \\ =\frac{\sin\theta+\sin\theta\cos\theta}{1-\cos^2\theta} \end{gathered}[/tex]We know the identity:
[tex]\begin{gathered} \sin ^2\theta+\cos ^2\theta=1 \\ or \\ \sin ^2\theta=1-\cos ^2\theta \end{gathered}[/tex]Substituting, we get:
[tex]\begin{gathered} \frac{\sin\theta+\sin\theta\cos\theta}{1-\cos^2\theta} \\ =\frac{\sin\theta+\sin\theta\cos\theta}{\sin^2\theta} \\ =\frac{\sin\theta}{\sin^2\theta}+\frac{\sin\theta\cos\theta}{\sin^2\theta} \\ =\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta} \end{gathered}[/tex]We know:
[tex]\frac{1}{\sin\theta}=\csc \theta[/tex]and
[tex]\begin{gathered} \frac{\sin\theta}{\cos\theta}=\tan \theta \\ \text{and} \\ \frac{\cos\theta}{\sin\theta}=\frac{1}{\tan\theta}=\cot \theta \end{gathered}[/tex]Of couse, we can see that it is proved.
[tex]\begin{gathered} \frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta} \\ =\csc \theta+\cot \theta \end{gathered}[/tex]