Kristin Montgomery

2023-01-01

How to prove $\frac{{\mathrm{tan}}^{2}x}{\mathrm{sec}x-1}=\mathrm{sec}x+1$?

Elgari8lx

Expert

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

yegumbi4q0

Expert

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.

Do you have a similar question?