Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.4 Trigonometric Identities - 6.4 Assess Your Understanding - Page 497: 40


The righ side of the equation is eqiuvalent to $\csc{u} - \cot{u}$ therefore the given equation is an identity. Refer to the solution below,

Work Step by Step

Work on the right hand side (RHS) of the equation. Multiply the RHS by $\dfrac{1-\cos{u}}{1-\cos{u}}$ to obtain: \begin{align} \text{RHS } &= \dfrac{\sin{u}}{1+\cos{u}} \times \dfrac{1-\cos{u}}{1-\cos{u}} \\[3mm] &= \dfrac{\sin{u}(1-\cos{u})}{1-\cos^2{u}} \\[3mm] &= \dfrac{\sin{u}(1-\cos{u})}{\sin^2{u}} \\[3mm] &= \dfrac{1-\cos{u}}{\sin{u}} \\[3mm] &= \dfrac{1}{\sin{u}}-\dfrac{\cos{u}}{\sin{u}}\\[3mm] &= \csc{u}-\cot{u} \\[3mm] &= \text{ LHS} \end{align}
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