Prove the following statement:
 $\left( \operatorname{cosec}A+\cot A \right)\operatorname{covers}A-\left( \sec A+\tan A \right)\operatorname{vers}A=\left( \operatorname{cosec}A-\sec A \right)\left( 2-\operatorname{vers}A\operatorname{covers}A \right)$.

Hint: In order to prove this given relation, we need to know that $\operatorname{vers}A=1-\cos A$ and $\operatorname{covers}A=1-\sin A$. Also, we should know that $\cot A=\dfrac{\cos A}{\sin A},\tan A=\dfrac{\sin A}{\cos A},\sec A=\dfrac{1}{\cos A}$ and $\operatorname{cosec}A=\dfrac{1}{\sin A}$. By using these properties, we can prove the desired relation.
Complete step-by-step answer:
In this question, we have to prove that, $\left( \operatorname{cosec}A+\cot A \right)\operatorname{covers}A-\left( \sec A+\tan A \right)\operatorname{vers}A=\left( \operatorname{cosec}A-\sec A \right)\left( 2-\operatorname{vers}A\operatorname{covers}A \right)$. For doing that, we will first consider the left hand side or the LHS of the equation, which is,
$LHS=\left( \operatorname{cosec}A+\cot A \right)\operatorname{covers}A-\left( \sec A+\tan A \right)\operatorname{vers}A$
We know that $\operatorname{covers}A=1-\sin A$ and $\operatorname{vers}A=1-\cos A$. So, substituting it, we get LHS as,
$LHS=\left( \operatorname{cosec}A+\cot A \right)\left( 1-\sin A \right)-\left( \sec A+\tan A \right)\left( 1-\cos A \right)$
Now, we know that, $\operatorname{cosec}A=\dfrac{1}{\sin A}$, $\cot A=\dfrac{\cos A}{\sin A},\sec A=\dfrac{1}{\cos A}$ and $\tan A=\dfrac{\sin A}{\cos A}$. So, by substituting them in the above equation, we get the LHS as,
$LHS=\left( \dfrac{1}{\sin A}+\dfrac{\cos A}{\sin A} \right)\left( 1-\sin A \right)-\left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A} \right)\left( 1-\cos A \right)$
We can further simplify and write it as,
$LHS=\left( \dfrac{1+\cos A}{\sin A} \right)\left( 1-\sin A \right)-\left( \dfrac{1+\sin A}{\cos A} \right)\left( 1-\cos A \right)$
Now, we will take the LCM of both the terms. So, we get,
$LHS=\dfrac{\left( 1+\cos A \right)\left( 1-\sin A \right)\left( \cos A \right)-\left( 1+\sin A \right)\left( 1-\cos A \right)\sin A}{\sin A\cos A}$
Now, we will open each bracket to simplify further, so we get LHS as,
$\begin{align}
  & LHS=\dfrac{\left( 1+\cos A-\sin A-\sin A\cos A \right)\left( \cos A \right)-\left( 1+\sin A-\cos A-\sin A\cos A \right)\left( \sin A \right)}{\sin A\cos A} \\
 & \Rightarrow LHS=\dfrac{\left( \cos A+{{\cos }^{2}}A-\sin A\cos A-\sin A{{\cos }^{2}}A \right)-\left( \sin A+{{\sin }^{2}}A-\sin A\cos A-{{\sin }^{2}}A\cos A \right)}{\sin A\cos A} \\
\end{align}$
We can further simplify it as,
$LHS=\dfrac{\cos A+{{\cos }^{2}}A-\sin A\cos A-\sin A{{\cos }^{2}}A-\sin A-{{\sin }^{2}}A+\sin A\cos A+{{\sin }^{2}}A\cos A}{\sin A\cos A}$
And we know that same terms of the opposite signs get cancelled, so we will cancel $\sin A\cos A$, and therefore we get the LHS as,
$LHS=\dfrac{\left( \cos A-\sin A \right)+{{\cos }^{2}}A-{{\sin }^{2}}A-\sin A{{\cos }^{2}}A+{{\sin }^{2}}A\cos A}{\sin A\cos A}$
Now, we can see that $\left( -\sin A{{\cos }^{2}}A+{{\sin }^{2}}A\cos A \right)$ can be written as $\left[ -\sin A\cos A\left( \cos A-\sin A \right) \right]$. And by using the identity, ${{a}^{2}}+{{b}^{2}}=\left( a-b \right)\left( a+b \right)$, we can write ${{\cos }^{2}}A-{{\sin }^{2}}A$ as $\left( \cos A-\sin A \right)\left( \cos A+\sin A \right)$. Therefore, we can write the LHS as,
$LHS=\dfrac{\left( \cos A-\sin A \right)+\left( \cos A-\sin A \right)\left( \cos A+\sin A \right)-\sin A\cos A\left( \cos A-\sin A \right)}{\sin A\cos A}$
Now, we can see that we can take out the common term, $\left( \cos A-\sin A \right)$ from the LHS, so we get,
$\begin{align}
  & LHS=\dfrac{\left( \cos A-\sin A \right)\left( 1+\cos A+\sin A-\cos A\sin A \right)}{\sin A\cos A} \\
 & \Rightarrow LHS=\dfrac{\cos A-\sin A}{\sin A\cos A}\left[ 1+\cos A+\sin A-\cos A\sin A \right] \\
\end{align}$
Now, we will add and subtract 1 from $\left[ 1+\cos A+\sin A-\cos A\sin A \right]$. So, we get the LHS as,
$\begin{align}
  & LHS=\dfrac{\cos A-\sin A}{\cos A\sin A}\left[ 1+\cos A+\sin A-\cos A\sin A+1-1 \right] \\
 & \Rightarrow LHS=\dfrac{\cos A-\sin A}{\cos A\sin A}\left[ 2+\cos A-\cos A\sin A-1+\sin A \right] \\
\end{align}$
Now we will take $\cos A$ common from $\left( \cos A-\cos A\sin A \right)$, and we get the LHS as,
$LHS=\dfrac{\cos A-\sin A}{\cos A\sin A}\left[ 2+\cos A\left( 1-\sin A \right)-1\left( 1-\sin A \right) \right]$
Now, we will take $\left[ -\left( 1-\sin A \right) \right]$ common from $\cos A\left( 1-\sin A \right)-1\left( 1-\sin A \right)$, so we get,
$LHS=\dfrac{\cos A-\sin A}{\cos A\sin A}\left[ 2-\left( 1-\sin A \right)\left( 1-\cos A \right) \right]$
And we know that $1-\sin A=\operatorname{covers}A$ and $1-\cos A=versA$. So, substituting it, we get LHS as,
$LHS=\dfrac{\cos A-\sin A}{\cos A\sin A}\left[ 2-\operatorname{vers}A\operatorname{covers}A \right]$
And we can further write it as,
$LHS=\left[ \dfrac{\cos A}{\cos A\sin A}-\dfrac{\sin A}{\cos A\sin A} \right]\left[ 2-\operatorname{vers}A\operatorname{covers}A \right]$
Cancelling the common terms from the numerator and denominator, we get,
$LHS=\left[ \dfrac{1}{\sin A}-\dfrac{1}{\cos A} \right]\left[ 2-\operatorname{vers}A\operatorname{covers}A \right]$
And we know that $\dfrac{1}{\sin A}=\operatorname{cosec}A$ and $\dfrac{1}{\cos A}=\sec A$, so substituting it, we get,
$LHS=\left[ \operatorname{cosec}A-\sec A \right]\left[ 2-\operatorname{vers}A\operatorname{covers}A \right]$
Which is equal to the right hand side of the given statement, therefore LHS = RHS.
Hence, we have proved the given statement.

Note: In this question, there are 2 points where we may get confused. The first point is where we wrote, $\operatorname{vers}A=1-\cos A$ and $\operatorname{covers}A=1-\sin A$ because these are the two most important formulas that are the requirements of this question. Also, some of us may get confused at the point where we added and subtracted 1, because sometimes in a hurry, we forget these points and in such a case, we can start from RHS, after that step.