Question

Explain why, if $x=asecu$, then $\sqrt{a^{2}se{c}^{2}u-a^2}$is $-a\tan u$if $x<-a$and is $a\tan u$if $x>a$. Your explanation should include a discussion of domains and absolute values.

Short answer

Ans:

$$\begin{gathered} \mathrm{The}\mathrm{absolute}\mathrm{value}\mathrm{sre}\mathrm{significant}\mathrm{since}\mathrm{tanu}\mathrm{is}\mathrm{positive}\mathrm{for}\mathrm{u}\mathrm{in}{1}^{\mathrm{st}}\mathrm{quadrant} \\ \mathrm{and}\mathrm{in}{2}^{\mathrm{st}}\mathrm{quadrant}\mathrm{its}\mathrm{negative}\mathrm{for}\mathrm{u}. \end{gathered}$$

Step-by-step solution

Step 1. Given information:

$$\begin{gathered} \sqrt{a^{2}se{c}^{2}u-a^2} \\ x=asecu \\ \left\{\begin{array}{ll}-a\tan u& x<-a\\ a\tan u& x>a\end{array}\right. \end{gathered}$$

Step 2. Solving the trigonometric substitution:

Suppose $x\in (-\infty ,a]\cup [a,\infty ),u\in \left[0,\frac{\pi }{2}\right)\cup \left(\frac{\pi }{2},\pi \right]$, and $x=asecu$

$$\begin{gathered} \sqrt{x^2-a^2} \\ =\sqrt{a^{2}se{c}^{2}u-a^2} \\ =\sqrt{a^2{\tan }^{2}u} \\ =\left|a\tan u\right| \\ \left\{\begin{array}{ll}-a\tan u& ,ifx<-a\\ a\tan u& ,ifx>a\end{array}\right. \\ Theabsolutevaluesresignificant\sin ce\tan uispositiveforuin{1}^{st}quadrant \\ andin{2}^{st}quadrantitsnegativeforu. \end{gathered}$$