Trigonometry 7th Edition

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

Chapter 5 - Section 5.3 - Double-Angle Formulas - 5.3 Problem Set - Page 298: 59

Answer

See the steps.

Work Step by Step

$LHS =\dfrac{1-\tan{x}}{1+\tan{x}} =\dfrac{1-\dfrac{\sin{x}}{\cos{x}}}{1+\dfrac{\sin{x}}{\cos{x}}}$ $LHS = \dfrac{\cos{x} -\sin{x}}{\cos{x} +\sin{x}}$ Multiply by $\dfrac{\cos{x} -\sin{x}}{\cos{x} -\sin{x}}$ $LHS =\dfrac{\cos^2{x}-2\sin{x}\cos{x}+\sin^2{x}}{\cos^2{x}-\sin^2{x}}$ $LHS =\dfrac{1-\sin{2x}}{\cos{2x}}= RHS$
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