# How do you use trigonometric substitution to write the algebraic expression sqrt(x^2-9) as a trigonometric function of theta where 0<theta<pi/2 and x=3sectheta?

Feb 13, 2017

The desired function is $3 \tan \theta$

#### Explanation:

You have already given hint for the solution.

If we put $x = 3 \sec \theta$, then $\sqrt{{x}^{2} - 9}$

= $\sqrt{9 {\sec}^{2} \theta - 9}$

= $\sqrt{9 \left({\sec}^{2} \theta - 1\right)}$

= $\sqrt{9 {\tan}^{2} \theta}$

= $3 \tan \theta$

Hence, the desired function is $3 \tan \theta$