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 280: 62

Answer

As left side transforms into right side, hence given identity- $ \frac{\sin^{2} B - \tan^{2} B}{1 - \sec^{2} B}$ = $\sin^{2} B $ is true.

Work Step by Step

Given identity is- $ \frac{\sin^{2} B - \tan^{2} B}{1 - \sec^{2} B}$ = $\sin^{2} B $ Taking L.S. $ \frac{\sin^{2} B - \tan^{2} B}{1 - \sec^{2} B}$ = $ \frac{-1(\tan^{2} B - \sin^{2} B )}{-1(\sec^{2} B - 1)}$ (taking (-1) out as common factor to change sign to suite trigonometric identities) = $ \frac{\tan^{2} B - \sin^{2} B }{\sec^{2} B - 1}$ {(-1) cancelled out as common factor} = $ \frac{\tan^{2} B - \sin^{2} B}{\tan^{2} B}$ ( From second Pythagorean identity, $\sec^{2} B - 1= \tan^{2} B$) = $ \frac{\tan^{2} B }{\tan^{2} B} - \frac{\sin^{2} B}{\tan^{2} B}$ = $ 1- \sin^{2} B \frac{1}{\tan^{2} B}$ = $ 1- \sin^{2} B \cot^{2} B$ = $ 1- \sin^{2} B \frac{\cos^{2} B}{\sin^{2} B}$ = $ 1- \cos^{2} B $ = $\sin^{2} B $ = R.S. ( From first Pythagorean identity, $1 -\cos^{2}B$ can be replaced with, $\sin^{2}B$) As left side transforms into right side, hence given identity- $ \frac{\sin^{2} B - \tan^{2} B}{1 - \sec^{2} B}$ = $\sin^{2} B $ is true.
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