Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.4 - Graphs of Polar Equations - Exercise Set - Page 755: 96


See the proof below.

Work Step by Step

Consider the left side of the equation $\frac{1+\sin x}{1-\sin x}-\frac{1-\sin x}{1+\sin x}$. At first, take the least common multiple of the denominator, $\begin{align} & =\frac{\left( 1+\sin x \right)\left( 1+\sin x \right)-\left( 1-\sin x \right)\left( 1-\sin x \right)}{\left( 1-\sin x \right)\times \left( 1+\sin x \right)} \\ & =\frac{\left( 1+{{\sin }^{2}}x+2\sin x \right)-\left( 1+{{\sin }^{2}}x-2\sin x \right)}{\left( 1-{{\sin }^{2}}x \right)} \\ \end{align}$ Then reduce the above equation in simplified form $\begin{align} & =4\frac{\sin x}{{{\cos }^{2}}x} \\ & =4\frac{\sin x}{\cos x\times \cos x} \\ \end{align}$ Also, $\frac{\sin x}{\cos x}=\tan x\,\text{ and }\,\frac{1}{\cos x}=\sec x $ The simplified form of equation is: $\begin{align} & =4\frac{\sin x}{\cos x}\times \frac{1}{\cos x} \\ & =4\tan x\sec x \\ \end{align}$ Hence, the identity is verified.
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