How to prove tan2xsec⁡x−1=sec⁡x+1?

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Elgari8lx · Accepted Answer

To prove tan2(x)sec⁡(x)−1=sec⁡(x)+1Use the identity 1+tan2(x)=sec2(x)This can be rewritten astan2(x)=sec2(x)−1returning to our issueLHS =tan2(x)sec⁡(x)−1=sec2(x)−1sec⁡(x)−1Remember the square-difference rule.a2−b2=(a−b)(a+b)We need to apply that for sec2(x)−1=(sec⁡(x)−1)(sec⁡(x)+1)sec⁡(x)−1Undefined control sequence cancelUndefined control sequence cancel=sec⁡(x)+1=RHSHence, LHS = RHS thus proved.

yegumbi4q0 · Accepted Answer

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.

How to prove tan^2x / (secx - 1) = secx + 1?

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