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: 60

Answer

See the steps.

Work Step by Step

$RHS = \sec{x} \csc{x} -\cot{x} +tan{x} = \dfrac{1}{\sin{x}\cos{x}} -\dfrac{\sin{x}}{\cos{x}}+\dfrac{\cos{x}}{\sin{x}}$ $RHS =\dfrac{1}{\sin{x}\cos{x}} - \dfrac{\sin^2{x}-\cos^2{x}}{\cos{x} \sin{x}}$ $RHS =\dfrac{2}{\sin{2x}} - \dfrac{2\cos{2x}}{\sin{2x}} = \dfrac{2-2\cos{2x}}{\sin{2x}} = LHS$
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