How to prove tan2xsec⁡x−1=sec⁡x+1?

To prove ${\displaystyle \frac{{\tan }^2\left(x\right)}{\sec \left(x\right)-1}=\sec \left(x\right)+1}$
Use the identity ${\displaystyle 1+{\tan }^2\left(x\right)={\sec }^2\left(x\right)}$
This can be rewritten as
${\displaystyle {\tan }^2\left(x\right)={\sec }^2\left(x\right)-1}$
returning to our issue
LHS
${\displaystyle =\frac{{\tan }^2\left(x\right)}{\sec \left(x\right)-1}}$
${\displaystyle =\frac{{\sec }^2\left(x\right)-1}{\sec \left(x\right)-1}}$
Remember the square-difference rule.
${\displaystyle a^2-b^2=(a-b)(a+b)}$
We need to apply that for ${\displaystyle {\sec }^2\left(x\right)-1}$
${\displaystyle =\frac{(\sec \left(x\right)-1)(\sec \left(x\right)+1)}{\sec \left(x\right)-1}}$
$\text{Undefined control sequence cancel}$
${\displaystyle =\sec \left(x\right)+1=}$RHS
Hence, LHS = RHS thus proved.

Choose the aspect that will be the most challenging to work on first. In this case, it's the left side. Recall the Pythagorean trigonometric identity, \(\displaystyle{\color{red}{{{\tan}^{{2}}{x}}}}={\color{green}{{{\sec}^{{2}}{x}}-{1}}}\). Using this identity, replace \(\displaystyle{\color{red}{{{\tan}^{{2}}{x}}}}\) in the equation with \(\displaystyle{\color{green}{{{\sec}^{{2}}{x}}-{1}}}\).
Left side:
\(\displaystyle\frac{{\color{red}{{{\tan}^{{2}}{x}}}}}{{{\sec{{x}}}-{1}}}\)
\(\displaystyle\frac{{{\color{green}{{{\sec}^{{2}}{x}}-{1}}}}}{{{\sec{{x}}}-{1}}}\)
Since ""\(\displaystyle{{\sec}^{{2}}{x}}-{1}\)"" is a difference of squares, it can be broken down into \(\displaystyle{\color{\quad\textor\quadan\ge}{{\sec{{x}}}+{1}}}\) and \(\displaystyle{\color{blue}{{\sec{{x}}}-{1}}}\).
\(\displaystyle\frac{{{\left({\color{\quad\textor\quadan\ge}{{\sec{{x}}}+{1}}}\right)}{\left({\color{blue}{{\sec{{x}}}-{1}}}\right)}}}{{{\sec{{x}}}-{1}}}\)
You will notice that ""\(\displaystyle{\sec{{x}}}-{1}\)"" appears both in the numerator and denominator, so they can both be cancelled out.
\(\displaystyle\frac{{{\left({\sec{{x}}}+{1}\right)}{\color{red}\cancel{{\color{black}{{\left({\sec{{x}}}-{1}\right)}}}}}}}{{\color{red}\cancel{{\color{black}{{\left({\sec{{x}}}-{1}\right)}}}}}}\)
\(\displaystyle{\sec{{x}}}+{1}\)
\(\displaystyle\therefore\), LS\(\displaystyle=\)RS.