Answer
See the full explanation below.
Work Step by Step
$\frac{\sin \theta }{1-\cot \theta }-\frac{\cos \theta }{\tan \theta -1}$
Apply the quotient identity of trigonometry $\cot \theta =\frac{\cos \theta }{\sin \theta }$ and $\tan \theta =\frac{\sin \theta }{\cos \theta }$. Then, the above expression can be further simplified as:
$\frac{\sin \theta }{1-\cot \theta }-\frac{\cos \theta }{\tan \theta -1}=\frac{\sin \theta }{1-\frac{\cos \theta }{\sin \theta }}-\frac{\cos \theta }{\frac{\sin \theta }{\cos \theta }-1}$
In the above expression, the least common denominator is $\sin \theta $ and $\cos \theta $, respectively. Rewrite each fraction with the least common denominator:
$\frac{\sin \theta }{1-\frac{\cos \theta }{\sin \theta }}-\frac{\cos \theta }{\frac{\sin \theta }{\cos \theta }-1}=\frac{\sin \theta }{\frac{\sin \theta -\cos \theta }{\sin \theta }}-\frac{\cos \theta }{\frac{\sin \theta -\cos \theta }{\cos \theta }}$
Then, rewrite the main fraction bar as $\div $:
$\frac{\sin \theta }{\frac{\sin \theta -\cos \theta }{\sin \theta }}-\frac{\cos \theta }{\frac{\sin \theta -\cos \theta }{\cos \theta }}=\sin \theta \div \frac{\sin \theta -\cos \theta }{\sin \theta }-\cos \theta \div \frac{\sin \theta -\cos \theta }{\cos \theta }$
Invert the divisor and multiply:
$\begin{align}
& \sin \theta \div \frac{\sin \theta -\cos \theta }{\sin \theta }-\cos \theta \div \frac{\sin \theta -\cos \theta }{\cos \theta }=\sin \theta .\frac{\sin \theta }{\sin \theta -\cos \theta }-\cos \theta .\frac{\cos \theta }{\sin \theta -\cos \theta } \\
& =\frac{{{\sin }^{2}}\theta }{\sin \theta -\cos \theta }-\frac{{{\cos }^{2}}\theta }{\sin \theta -\cos \theta }
\end{align}$
By expressing the terms in the numerator and denominator with the least common denominator $\sin \theta -\cos \theta $, we get:
$\frac{{{\sin }^{2}}\theta }{\sin \theta -\cos \theta }-\frac{{{\cos }^{2}}\theta }{\sin \theta -\cos \theta }=\frac{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta }{\sin \theta -\cos \theta }$
We use the formulae ${{A}^{2}}-{{B}^{2}}=(A+B)(A-B)$, with $A={{\sin }^{2}}\theta $ and $B={{\cos }^{2}}\theta $. Now, rewrite the expression as:
$\begin{align}
& \frac{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta }{\sin \theta -\cos \theta }=\frac{\left( \sin \theta +\cos \theta \right)\left( \sin \theta -\cos \theta \right)}{\sin \theta -\cos \theta } \\
& =\sin \theta +\cos \theta
\end{align}$
Thus, the left side of the expression is equal to the right side, which is
$\frac{\sin \theta }{1-\cot \theta }-\frac{\cos \theta }{\tan \theta -1}=\sin \theta +\cos \theta $.
