Kristin Montgomery

2023-01-01

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

Elgari8lx

Beginner2023-01-02Added 7 answers

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.

${a}^{2}-{b}^{2}=(a-b)(a+b)$

We need to apply that for${\mathrm{sec}}^{2}\left(x\right)-1$

$=\frac{(\mathrm{sec}\left(x\right)-1)(\mathrm{sec}\left(x\right)+1)}{\mathrm{sec}\left(x\right)-1}$

$\text{Undefined control sequence cancel}$

$=\mathrm{sec}\left(x\right)+1=$ RHS

Hence, LHS = RHS thus proved.

Use the identity

This can be rewritten as

returning to our issue

LHS

Remember the square-difference rule.

We need to apply that for

Hence, LHS = RHS thus proved.

yegumbi4q0

Beginner2023-01-03Added 3 answers

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.

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.

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