Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 680: 68


See the explanation below.

Work Step by Step

Let us consider the left side of the given expression: $\tan \frac{x}{2}-\cot \frac{x}{2}$ By using the trigonometric identity $\tan \frac{x}{2}=\frac{1-\cos x}{\sin x}$ and $\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}$, the above expression can be further simplified as: $\begin{align} & \tan \frac{x}{2}-\cot \frac{x}{2}=\frac{1-\cos x}{\sin x}-\frac{1}{\tan \frac{x}{2}} \\ & =\frac{1-\cos x}{\sin x}-\frac{1}{\frac{\sin x}{1+\cos x}} \\ & =\frac{1-\cos x}{\sin x}-\frac{\left( 1+\cos x \right)}{\sin x} \end{align}$ Now, taking the LCM, the equation will be $\begin{align} & \frac{1-\cos x}{\sin x}-\frac{\left( 1+\cos x \right)}{\sin x}=\frac{1-\cos x-1-\cos x}{\sin x} \\ & =\frac{-2\cos x}{\sin x} \\ & =-2\frac{\cos x}{\sin x} \\ & =-2\cot x \end{align}$ Hence, the left side of the given expression is equal to the right side, which is $\tan \frac{x}{2}-\cot \frac{x}{2}=-2\cot x$.
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