Prove the given expression: $\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \cos ec\theta $

Hint: Start with taking LCM in LHS and simplify it using the identity ${{a}^{3}}-{{b}^{3}}$ , Now convert tan into sin and cos then use the fact that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ for all $\theta $ to get the required answer.

Complete step-by-step answer:

An equality with sine, cosine or tangent in them is called a trigonometric equation. These are solved by interrelations known beforehand.
All the interrelations which relate sine, cosine, tangent, secant, cotangent, cosecant are called trigonometric identities. These trigonometric identities solve the equation and make them simpler to understand for a proof. These are the main and crucial steps to take us nearer to result.
Given equation in the question which we need to prove:
$\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \cos ec\theta $
Take the left hand separately and simplify it
$\dfrac{\tan \theta }{1-\tan \theta }+\dfrac{\dfrac{1}{\tan \theta }}{1-\tan \theta }$
By taking the common term outside the expression we can turn into:
$\dfrac{1}{\tan \theta -1}\left( {{\tan }^{2}}\theta -\dfrac{1}{\tan \theta } \right)=\dfrac{{{\tan }^{3}}\theta -1}{\tan \theta \left( \tan \theta -1 \right)}$
By simplifying further, the expression turns into form of:
$\dfrac{\left( \tan \theta -1 \right)\left( {{\tan }^{2}}\theta +1+\tan \theta \right)}{\tan \theta \left( \tan \theta -1 \right)}$
By cancelling the term and using relation between tan, cot, we get
$\tan \theta +\cot \theta +1$
By converting all the terms into sine, cosines, we get that:
$\dfrac{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}{\sin \theta \cos \theta }+1=1+\sec \theta \cos ec\theta $ by using the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ we proved the required equation.
Hence proved.

Note: (i) Be careful while using the algebraic identity ${{a}^{3}}-{{b}^{3}}$
(ii) After converting into sine, cosine takes the L.C.M properly.
(iii) You can directly take LCM in the first step and solve in a brute force method. But it will take some time to get results.