In the given question prove that $(\sec A + \tan A - 1)(\sec A - \tan A + 1) - 2\tan A = 0$

Hint: Use appropriate trigonometric identities and solve it, in this case, use the trigonometry identity ${\sec ^2}\theta - {\tan ^2}\theta = 1$ and solve it.

Complete Step-by-Step solution:
Here we have to prove that L.H.S=R.H.S
So, let us take the L.H.S part
\[(\sec A + \tan A - 1)(\sec A - \tan A + 1) - 2\tan A\]
Consider $a = \sec A$ and $b = \tan A + 1$
We know that $(a + b)(a - b) = {a^2} - {b^2}$
On applying the $(a + b)(a - b) = {a^2} - {b^2}$ formula we can rewrite the L.H.S as
$ \Rightarrow ({\sec ^2}A) - {(\tan A - 1)^2} - 2\tan A$
Apply ${(a - b)^2} = {a^2} + {b^2} - 2ab$ on ${(\tan A - 1)^2}$ then we get
$
   \Rightarrow {\sec ^2}A - {\tan ^2}A - 1 + 2\tan A - 2\tan A \\
   \Rightarrow {\sec ^2}A - {\tan ^2}A - 1 \\
$
Apply the trigonometry identity ${\sec ^2}\theta - {\tan ^2}\theta = 1$ in above equation we get
$ \Rightarrow 1 - 1 = 0$
$ \Rightarrow $R.H.S
Hence we have proved that L.H.S=R.H.S.

NOTE: In these types of problems, make use of appropriate trigonometric identities and modify the LHS in accordance to the RHS which has to be proved.