Q. 13 Fill in the blanks to complete e... [FREE SOLUTION]

Chapter 5: Q. 13 (page 494)

Fill in the blanks to complete each of the following theorem statements:
If $a$ is a positive real number, and if $x 鈭�___$ and role="math" localid="1653539365259" $u 鈭�_$ , then the substitution $x = a s e c u$ gives us $\sqrt{x^{2} - a^{2}} =_$ if $x < - a$ and $\sqrt{x^{2} - a^{2}} =___$ if $x > a$ .

Short Answer

Expert verified

If $a$ is a positive real number, and if $x 鈭� (鈭� 鈭�, 鈭� a] 鈭� [a, 鈭�)$ and $u 鈭� [0, \frac{蟺}{2}) 鈭� (\frac{蟺}{2}, 蟺]$ , then the substitution $x = a s e c u$ gives us $\sqrt{x^{2} - a^{2}} = - a \tan u$ if $x < - a$ and $\sqrt{x^{2} - a^{2}} = a \tan u$ if $x > a$ .

Step by step solution

Step 1. Given information

If $a$ is a positive real number, and if $x 鈭�_____$ and $u 鈭�_____$ , then the substitution $x = a s e c u$ gives us $\sqrt{x^{2} - a^{2}} =_____$ if $x < - a$ and $\sqrt{x^{2} - a^{2}} =_____$ if $x > a$ .

Step 2. Filling in the blanks to complete the theorem statements

By substituting for $x$ and simplifying, we get the above answer valid under the conditions mentioned. That is, role="math" localid="1653540018880" $\sqrt{x^{2} - a^{2}} = \sqrt{{(a s e c u)}^{2} - a^{2}}$

role="math" localid="1653540023117" $= \sqrt{a^{2} (s e c^{2} u - 1)} = \sqrt{a^{2} \tan^{2} u} = \{\begin{cases} - a \tan u, i f x < - a \\ a \tan u, i f x > a \end{cases}$

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Short Answer

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Step 1. Given information

Step 2. Filling in the blanks to complete the theorem statements

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Fill in the blanks to complete each of the following theorem statements: If ais a positive real number, and if x鈭�_____and role="math" localid="1653539365259" u鈭�_____, then the substitution x=asecugives us x2-a2=_____if x<-aand x2-a2=_____if x>a.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Filling in the blanks to complete the theorem statements

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Discrete Mathematics

Mechanics Maths

Logic and Functions

Pure Maths

Calculus

Study anywhere. Anytime. Across all devices.