Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - Chapter 7 Test Prep - Review Exercises - Page 737: 56


$\frac{2\cot x}{\tan 2x}=\csc^2 x-2$

Work Step by Step

Start with the left side: $\frac{2\cot x}{\tan 2x}$ Simplify using the Double-Angle Formula for tangent: $=\frac{2\cot x}{\frac{2\tan x}{1-\tan^2 x}}$ Multiply the top and bottom by $1-\tan^2 x$: $=\frac{2\cot x}{\frac{2\tan x}{1-\tan^2 x}}*\frac{1-\tan^2 x}{1-\tan^2 x}$ Simplify: $=\frac{2\cot x(1-\tan^2 x)}{2\tan x}$ $=\frac{\cot x(1-\tan^2 x)}{\frac{1}{\cot x}}$ Multiply the top and bottom by $\cot x$: $=\frac{\cot x(1-\tan^2 x)}{\frac{1}{\cot x}}*\frac{\cot x}{\cot x}$ Simplify: $=\cot^2 x(1-\tan^2 x)$ $=\cot^2 x-1$ $=(\csc^2 x-1)-1$ $=\csc^2 x-2$ Since this equals the right side, the identity has been proven.
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