Chapter 5: Q. 15 (page 464)

Explain why, if $x = a s e c u$ , then $\sqrt{a^{2} s e c^{2} u - a^{2}}$ 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

Expert verified

Ans:

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

Step by step solution

Step 1. Given information:

$\sqrt{a^{2} s e c^{2} u - a^{2}} x = a s e c u \{\begin{cases} - a \tan u & x < - a \\ a \tan u & x > a \end{cases}$

Step 2. Solving the trigonometric substitution:

Suppose $x \in (- \infty, a] \cup [a, \infty), u \in [0, \frac{π}{2}) \cup (\frac{π}{2}, π]$ , and $x = a s e c u$

$\sqrt{x^{2} - a^{2}} = \sqrt{a^{2} s e c^{2} u - a^{2}} = \sqrt{a^{2} \tan^{2} u} = |a \tan u| \{\begin{cases} - a \tan u & , i f x < - a \\ a \tan u & , i f x > a \end{cases} T h e a b s o l u t e v a l u e s r e s i g n i f i c a n t \sin c e \tan u i s p o s i t i v e f o r u i n 1^{s t} q u a d r a n t a n d i n 2^{s t} q u a d r a n t i t s n e g a t i v e f o r u .$

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Explain why, if $x = a s e c u$ , then $\sqrt{a^{2} s e c^{2} u - a^{2}}$ 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

Step by step solution

Step 1. Given information:

Step 2. Solving the trigonometric substitution:

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Explain why, if x=asecu, then a2sec2u−a2is −atanuif x<−aand is atanuif x>a. Your explanation should include a discussion of domains and absolute values.

Short Answer

Step by step solution

Step 1. Given information:

Step 2. Solving the trigonometric substitution:

One App. One Place for Learning.

Most popular questions from this chapter

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Explain why, if $x = a s e c u$ , then $\sqrt{a^{2} s e c^{2} u - a^{2}}$ 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.