# Thread: Need help simplifying this trigonometric substitution problem on homework.

1. ## Need help simplifying this trigonometric substitution problem on homework.

I need help simplifying (sqrt(x^2-49))/(x) where x =7sec(theta)

I'm completely stumped, Trigonometry has proven to not be my strongest subject. Any help, or something worked out to push me in the right direction, would be fantastic!

2. ## Re: Need help simplifying this trigonometric substitution problem on homework.

sqrt(x^2 - 49)/x where x = 7 sec(theta)
sqrt(7(sec(theta)))^2 - 49) / (7sec(theta))
sqrt(49(sec(theta))^2 - 49)/ (7sec(theta))
sqrt(49[(sec(theta))^2 - 1]) / (7sec(theta))
sqrt{49(tan(theta))^2 / (7sec(theta))
(7 tan(theta)) / (7sec(theta))
7 tan(theta) cos(theta)/ 7
tan(theta) cos(theta)
sin(theta)/cos(theta) * cos(theta)
sin(theta)

3. ## Re: Need help simplifying this trigonometric substitution problem on homework.

Hello, AKPunk49!

I'll write it out in LaTeX . . . may be easier to read.

$\displaystyle \text{Simplify: }\:\frac{\sqrt{x^2-49}}{x},\:\text{ where }x = 7\sec\theta$

$\displaystyle \frac{\sqrt{x^2-49}}{x} \;=\;\frac{\sqrt{(7\sec\theta)^2 - 49}}{7\sec\theta} \;=\;\frac{\sqrt{49\sec^2\!\theta - 49}}{7\sec\theta}$

. . . . . . . $\displaystyle =\;\frac{\sqrt{49(\sec^2\!\theta-1)}}{\7\sec\theta} \;=\; \frac{\sqrt{49\tan^2\!\theta}}{7\sec\theta}$

. . . . . . . $\displaystyle =\;\frac{7\tan\theta}{7\sec\theta} \;=\;\frac{\tan\theta}{\sec\theta} \;=\;\frac{\frac{\sin\theta}{\cos\theta}}{\frac{1} {\cos\theta}} \;=\;\sin\theta$