Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 5 - Section 5.1 - Proving Identities - 5.1 Problem Set - Page 279: 13

Answer

13 (a). $ 2 \sqrt 2 - 3$ 13 (b). $(\sec x - \tan x)^{2}$

Work Step by Step

13 (a). Given expression is- $\frac{1 - \sqrt 2}{1 + \sqrt 2}$ = $\frac{1 - \sqrt 2}{1 + \sqrt 2}. \frac{1 - \sqrt 2}{1 - \sqrt 2}$ {Multiplying the numerator and denominator of the fraction by the conjugate of the denominator, $(1 - \sqrt 2)$} = $\frac{(1 - \sqrt 2)^{2}}{(1 - \sqrt 2)(1 +\sqrt 2)}$ = $\frac{1 - 2\sqrt 2 + 2}{{1^{2} - (\sqrt 2)^{2}}}$ {Recall $(a-b)^{2} = a^{2} -2ab +b^{2}$ and $(a-b)(a+b) $ = $a^{2} - b^{2}$ } = $\frac{ 3 - 2 \sqrt 2}{1 - 2}$ = $\frac{3 - 2 \sqrt 2}{-1}$ = $- (3 - 2 \sqrt 2)$ = $ 2 \sqrt 2 - 3$ 13 (b). Given expression is- $\frac{1 -\sin x}{1 + \sin x}$ = $\frac{1 -\sin x}{1 + \sin x}. \frac{1 -\sin x}{1 -\sin x}$ {Multiplying the numerator and denominator of the fraction by the conjugate of the denominator, $(1 -\sin x)$} = $\frac{(1 - \sin x)^{2}}{(1 + \sin x)(1 -\sin x)}$ = $\frac{1 - 2\sin x + \sin^{2}x}{{1^{2} - (\sin x)^{2}}}$ {Recall $(a-b)^{2} = a^{2} -2ab +b^{2}$ and $(a-b)(a+b) $ = $a^{2} - b^{2}$ } = $\frac{1 - 2\sin x + \sin^{2}x}{1 - \sin ^{2} x}$ = $\frac{1 - 2\sin x + \sin^{2}x}{\cos^{2} x}$ ( From first Pythagorean identity) = $\frac{1}{\cos^{2} x} - \frac{2 \sin x}{\cos^{2} x} + \frac{\sin^{2} x}{\cos^{2} x}$ = $\sec^{2} x - 2.\frac{1}{\cos x}.\frac{\sin x}{\cos x} + \tan^{2} x $ (using ratio and reciprocal identities) = $\sec^{2} x - 2.\sec x \tan x + \tan^{2} x $ = $(\sec x - \tan x)^{2}$
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