Prove the following statement: ${{\left( \tan \alpha +\csc \beta \right)}^{2}}-{{\left( \cot \beta -\sec \alpha \right)}^{2}}=2\tan \alpha \cot \beta \left( \csc \alpha +\sec \beta \right)$

Hint: We will start to solve the above expression from LHS, and we will use the formula of ${{\left( a+b \right)}^{2}}$ to expand the LHS and then we are going to prove the above statement by using some trigonometric formulas and then we will start from LHS and then by using the formulas we will show that it is equal to RHS.

Complete step-by-step answer:

Let’s first write all the formulas that are needed to solve this question.

Let’s first write trigonometric formula:

$\begin{align}


  & \cot A=\dfrac{\cos A}{\sin A} \\


 & \tan A=\dfrac{\sin A}{\cos A} \\


 & \sec A=\dfrac{1}{\cos A} \\


 & \csc A=\dfrac{1}{\sin A} \\


\end{align}$


Some more formula that we will use:


$\begin{align}


  & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\


 & {{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\


\end{align}$


Let’s first use this expansion formula to expand LHS,


$\begin{align}


  & {{\left( \tan \alpha +\csc \beta \right)}^{2}}-{{\left( \cot \beta -\sec \alpha \right)}^{2}} \\


 & ={{\tan }^{2}}\alpha +{{\csc }^{2}}\beta +2\tan \alpha \csc \beta -\left( {{\cot }^{2}}\beta
+{{\sec }^{2}}\alpha -2\cot \beta \sec \alpha \right) \\

\end{align}$


Now we will use these two formula:


$\begin{align}


  & {{\tan }^{2}}x-{{\sec }^{2}}x=-1 \\


 & {{\csc }^{2}}x-{{\cot }^{2}}x=1 \\


\end{align}$


Using these we get,


$\begin{align}


  & ={{\tan }^{2}}\alpha +{{\csc }^{2}}\beta +2\tan \alpha \csc \beta -{{\cot }^{2}}\beta -{{\sec }^{2}}\alpha +2\cot \beta \sec \alpha \\


 & =-1+1+2\tan \alpha \csc \beta +2\cot \beta \sec \alpha \\


 & =2\tan \alpha \left( \dfrac{\cos \beta }{\sin \beta \cos \beta } \right)+2\cot \beta \left( \dfrac{\sin \alpha }{\cos \alpha \sin \alpha } \right) \\


 & =2\tan \alpha \cot \beta \sec \beta +2\cot \beta \tan \alpha \csc \alpha \\


\end{align}$


Now taking the common terms out we get,


$2\tan \alpha \cot \beta \left( \csc \alpha +\sec \beta \right)$


Hence we have shown using the above formula that LHS = RHS.

Hence, proved.

Note: It’s always better that we check if the answer that we have got by using the above formula is correct or not to avoid some calculation mistake and for that we need to put some values in place of A to check whether it satisfies the above expression or not. There are a bunch of trigonometric formulas that should be kept in mind while solving these questions and if we use some different set of formulas then that will be another method to solve this question, but at some point we can see that they are nearly equal. There are many ways to solve this question as one can start from RHS and then using some of the trigonometric formula one can show that RHS = LHS.