# How do you use trigonometric substitution to write the algebraic expression sqrt(x^2+100) as a trigonometric function of theta where 0<theta<pi/2 and x=10tantheta?

May 19, 2018

$\left(H y p o t e\right) = \sqrt{{x}^{2} + 100}$

#### Explanation:

show below

$x = 10 \tan \theta$

$\tan \theta = \frac{x}{10}$

$\tan \theta = \frac{O p p o s i t e}{a \mathrm{dj} a c e n t}$

from Pythagorean theorem in perpendicular triangle

${\left(H y p o t e\right)}^{2} = {\left(O p p o s i t e\right)}^{2} + {\left(a \mathrm{dj} a c e n t\right)}^{2}$

${\left(H y p o t e\right)}^{2} = {\left(x\right)}^{2} + {\left(10\right)}^{2} = {x}^{2} + 100$

$\left(H y p o t e\right) = \sqrt{{x}^{2} + 100}$