Prove the following statement: $2ver\sin A+{{\cos }^{2}}A=1+ver{{\sin }^{2}}A$

Hint: First we are going to see what does version mean and then we are going to prove the above statement by using some trigonometric formulas and then we will start from RHS and then by using the formulas we will show that it is equal to LHS, but here we have to solve both RHS and LHS to convert them into simpler form.

Complete step-by-step solution:
So, let’s look at version,
$ver\sin A=1-\cos A$
Now using this formula in the above expression $2ver\sin A+{{\cos }^{2}}A=1+ver{{\sin }^{2}}A$ we get,
$\begin{align}
  & 2\left( 1-\cos A \right)+{{\cos }^{2}}A=1+{{\left( 1-\cos A \right)}^{2}} \\
 & 1+{{\cos }^{2}}A-2\cos A={{\left( 1-\cos A \right)}^{2}} \\
\end{align}$
Now we will use this formula for LHS,
${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Using the above formula we get LHS as,
$\begin{align}
  & {{\cos }^{2}}A-2\cos A+1 \\
 & ={{\left( 1-\cos A \right)}^{2}} \\
\end{align}$
Hence from this we have shown that LHS = RHS.
Hence we have proved the given statement.

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.