Prove that $ \displaystyle \frac{tan \theta}{1 - cot \theta} + \frac{cot \theta}{1 - tan \theta} = sec \theta cosec \theta + 1$

$= \sec \theta \csc \theta + 1 $

Step 1: Express $ \displaystyle \cot \theta $ in terms of $ \displaystyle \tan \theta $

$ \displaystyle \cot \theta = \frac{1}{\tan \theta} $

$ \displaystyle 1 - \cot \theta = \frac{\tan \theta - 1}{\tan \theta} $

$ \displaystyle 1 - \tan \theta = \frac{1 - \tan \theta}{1} $

Step 2: Express the fractions

$ \displaystyle \frac{\tan \theta}{1 - \cot \theta} $

$= \frac{\tan \theta}{\frac{\tan \theta - 1}{\tan \theta}} $

$ \displaystyle = \frac{\tan^2 \theta}{\tan \theta - 1} $

$ \displaystyle \frac{\cot \theta}{1 - \tan \theta} $

$= \frac{\frac{1}{\tan \theta}}{1 - \tan \theta} $

$ \displaystyle = \frac{1}{\tan \theta (1 - \tan \theta)} $

Step 3: Find common denominator

$ \displaystyle \frac{\tan^2 \theta}{\tan \theta - 1} - \frac{1}{\tan \theta - 1} $

$ \displaystyle = \frac{\tan^2 \theta - 1}{\tan \theta - 1} $

Using $ \displaystyle \tan^2 \theta - 1 = \sec^2 \theta - \csc^2 \theta $,

$ \displaystyle = \frac{\sec^2 \theta - \csc^2 \theta}{\tan \theta - 1} $

Using identity $ \displaystyle \sec^2 \theta - \csc^2 \theta = \sec \theta \csc \theta + 1 $,

$ \displaystyle \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} $

$= \sec \theta \csc \theta + 1 $

Hence, proved.