Prove that \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }\ =\ 1+\sec \theta \csc \theta \]by using formula and identities.

Hint: First consider left hand side of equation and then convert \[\tan \theta \] as \[\dfrac{\sin \theta }{\cos \theta }\] and \[\cot \theta \] as \[\dfrac{\cos \theta }{\sin \theta }\] and then do the further calculations to get the Right hand side of equation.

Complete step-by-step answer:
In the question we have to prove that
\[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }\ =\ 1+\sec \theta \csc \theta \]
Now, let’s consider the left had side of the equation we see that,
\[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }\] …………………………………………………………………………..(i)
Now we will use formula that is\[\tan \theta \ =\ \dfrac{\sin \theta }{\cos \theta }\] and \[\cot \theta \ =\ \dfrac{\cos \theta }{\sin \theta }\] , then substitute it in expression (i) so we get,
\[\begin{align}
  & \dfrac{\dfrac{\sin \theta }{\cos \theta }}{1-\dfrac{\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{1-\dfrac{\sin \theta }{\cos \theta }} \\
 & \Rightarrow \dfrac{\dfrac{\sin \theta }{\cos \theta }}{\dfrac{sin\theta -\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{\dfrac{cos\theta -\sin \theta }{\cos \theta }} \\
 & \Rightarrow \dfrac{{{\sin }^{2}}\theta }{cos\theta (\sin \theta -\cos \theta )}+\dfrac{{{\cos }^{2}}\theta }{\sin \theta (\cos \theta -\sin \theta )} \\
 & \Rightarrow \dfrac{{{\sin }^{2}}\theta }{\sin \theta \cos \theta -{{\cos }^{2}}\theta }+\dfrac{{{\cos }^{2}}\theta }{\sin \theta \cos \theta -{{\sin }^{2}}\theta }\ \\
\end{align}\]
Now we take LCM of the term so we get,
\[\dfrac{{{\sin }^{2}}\theta \left( \sin \theta \cos \theta -{{\sin }^{2}}\theta \right)+{{\cos }^{2}}\theta \left( \sin \theta \cos \theta -{{\cos }^{2}}\theta \right)}{\left( \sin \theta \cos \theta -{{\cos }^{2}}\theta \right)\left( \sin \theta \cos \theta -{{\sin }^{2}}\theta \right)}\]
So, it can be further simplified as,
\[\dfrac{{{\sin }^{3}}\theta \cos \theta -{{\sin }^{4}}\theta +\sin \theta {{\cos }^{3}}\theta -{{\cos }^{4}}\theta }{\cos \theta \left( \sin \theta -\cos \theta \right)\sin \theta \left( \cos \theta -\sin \theta \right)}\]
Which can write it as,
\[\dfrac{\sin \theta \cos \theta \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)-\left( {{\sin }^{4}}\theta +{{\cos }^{4}}\theta \right)}{-\sin \theta \cos \theta {{\left( \sin \theta -\cos \theta \right)}^{2}}}\]
Now, we will change \[\left( {{\sin }^{4}}\theta +{{\cos }^{4}}\theta \right)\] as \[\left\{ \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right\}\] and we will use identity \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta \ =\ 1\]. Hence, it can be written as,
\[\dfrac{\sin \theta \cos \theta -\left\{ 1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right\}}{-\sin \theta \cos \theta {{\left( \sin \theta -\cos \theta \right)}^{2}}}\]
Now, we will multiply -1 to both numerator and denominator so we get,
\[\dfrac{1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta -\sin \theta \cos \theta }{\sin \theta \cos \theta {{\left( \sin \theta -\cos \theta \right)}^{2}}}\]
Now, we will factorize the numerator so we will get,
\[\dfrac{1-2\sin \theta \cos \theta +\sin \theta \cos \theta -2{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{\sin \theta \cos \theta {{\left( \sin \theta -\cos \theta \right)}^{2}}}\]
Now, it can be written as,
\[\dfrac{\left( 1-2\sin \theta \cos \theta \right)\left( 1+\sin \theta \cos \theta \right)}{\sin \theta \cos \theta {{\left( \sin \theta -\cos \theta \right)}^{2}}}\]
Now, we will expand and write,
\[{{\left( \sin \theta -\cos \theta \right)}^{2}}\ =\ 1-2\sin \theta \cos \theta \]
So, we can write as,
\[\dfrac{\left( 1-2\sin \theta \cos \theta \right)\left( 1+\sin \theta \cos \theta \right)}{\sin \theta \cos \theta \left( 1-2\sin \theta \cos \theta \right)}\]
Hence, the fraction can be written as,
\[\dfrac{1+\sin \theta \cos \theta }{\sin \theta \cos \theta }\]
So, we will break into two fractions we get,
\[\dfrac{1}{\sin \theta \cos \theta }+\dfrac{\sin \theta \cos \theta }{\sin \theta \cos \theta }\]
\[=\ \dfrac{1}{\sin \theta \cos \theta }+1=\ 1+\sec \theta \csc \theta \]
Hence, \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }\]is equal to \[1+\sec \theta \csc \theta \] hence proved.

Note: Student should know identities such as \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta \ =\ 1\], formula such as \[\tan \theta \ =\ \dfrac{\sin \theta }{\cos \theta }\]and\[\cot \theta \ =\ \dfrac{\cos \theta }{\sin \theta }\].Making some necessary adjustments during simplification is key here.