Step 1: Convert everything to sine and cosine right from the start, instead of working with $\tan$ and $\cot$ algebra.
\[ \text{LHS} = \frac{\sin\theta/\cos\theta}{1-\cos\theta/\sin\theta} + \frac{\cos\theta/\sin\theta}{1-\sin\theta/\cos\theta} \]
Step 2: Simplify each of the two denominators.
\[ 1-\frac{\cos\theta}{\sin\theta}=\frac{\sin\theta-\cos\theta}{\sin\theta}, \qquad 1-\frac{\sin\theta}{\cos\theta}=\frac{\cos\theta-\sin\theta}{\cos\theta} \]
So the two fractions become $\frac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)}$ and $\frac{\cos^2\theta}{\sin\theta(\cos\theta-\sin\theta)}$.
Step 3: Bring them to a common denominator, noting the sign flip in the second term.
\[ \text{LHS} = \frac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)} - \frac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)} = \frac{\sin^3\theta-\cos^3\theta}{\sin\theta\cos\theta(\sin\theta-\cos\theta)} \]
Using $\sin^3\theta-\cos^3\theta=(\sin\theta-\cos\theta)(\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta)$, the $(\sin\theta-\cos\theta)$ cancels, leaving $\frac{\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta}{\sin\theta\cos\theta}$.
Step 4: Split the fraction and use $\sin^2\theta+\cos^2\theta=1$ to finish.
\[ \frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta} + \frac{\sin\theta\cos\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} + 1 \]
Since $\frac{1}{\sin\theta\cos\theta} = \frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta} = \tan\theta+\cot\theta$, we get $\text{LHS} = 1+\tan\theta+\cot\theta = \text{RHS}$, so the identity is proved.