How to prove tan2xsec⁡x−1=sec⁡x+1?

Kristin Montgomery

Kristin Montgomery

Answered

2023-01-01

How to prove tan2xsecx1=secx+1?

Answer & Explanation

Elgari8lx

Elgari8lx

Expert

2023-01-02Added 7 answers

To prove tan2(x)sec(x)1=sec(x)+1
Use the identity 1+tan2(x)=sec2(x)
This can be rewritten as
tan2(x)=sec2(x)1
returning to our issue
LHS
=tan2(x)sec(x)1
=sec2(x)1sec(x)1
Remember the square-difference rule.
a2b2=(ab)(a+b)
We need to apply that for sec2(x)1
=(sec(x)1)(sec(x)+1)sec(x)1
Undefined control sequence \cancel
=sec(x)+1=RHS
Hence, LHS = RHS thus proved.
yegumbi4q0

yegumbi4q0

Expert

2023-01-03Added 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.

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